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Translational Dynamics

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Why do things move? It seems such a profound question for something that happens around us all the time. In fact, it's something that puzzled scientists and philosophers for hundreds of years. For example, Aristotle believed that all objects had a 'natural' place to which objects fell towards depending on which 'element' they were made out of. However, it wasn't until Newton's laws of motion that an understanding of forces as the causes of motion was systematically laid out. In this article, we're going to look at translational dynamics which analyses forces and how they produce a type of motion known as translational motion. Before looking at translational dynamics, let's recap Newton's law of motion.

Newton's Laws of Motion

Newton's laws of motion are three fundamental laws that describe the forces acting on an object and its motion. Sir Isaac Newton published his principles of motion in Philosophiae Naturalis Principia Mathematica on July 5, 1687. These simple laws served as the foundation of classical mechanics, and Newton himself utilized them to describe a wide range of phenomena relating to the motion of physical objects.

Newton's First Law

Newton's first law is concerned with the idea of inertia. It states that an object will remain at rest or in uniform (i.e at constant velocity) motion in a straight line unless it is acted upon by a net external force. It states that objects don't just start moving, or change velocity, of their own accord, there must be a force involved. Alternatively, we can summarise the law as 'no acceleration means there is no force.

Newton's First Law - An object will remain at rest or in uniform motion in a straight line (i.e at constant velocity) unless it is acted upon by a net external force.

It's important to remember that because velocity is a vector with a direction, changing direction is a form of acceleration. This is why the law states that an object 'remains in a straight line unless acted upon by an external force. This idea prompted Newton to suggest that the planets orbiting the sun experience the force of gravity.

Fig. 1 - A box on the ground will stay stationary until an external force is applied. This is because the weight of the box and the normal force cancel each other, meaning the net force is zero.

If an object is stationary this doesn't necessarily mean that no forces are acting on it, simply that the sum of all the forces, the net force, is zero. A stationary box is affected by both its weight and the normal force of the floor pushing up on it. Because their vector sum equals zero, these two forces cancel each other out, and the net force is zero. That's why the box doesn't just fall through the floor!

Newton’s Second Law

Of all the three laws of motion, Newton's second law is the one that gives a direct mathematical relationship between the motion of an object and the force it experiences. It will probably be the single law of physics that you use most in all your studies.

Newton's Second Law - The resultant force acting on an object is equal to the product of the objects mass and its acceleration.

This is expressed mathematically by the famous equation:

$F=ma$

where $$F$$ is the resultant or net force measured in Newtons $$\mathrm{N}$$, $$m$$ is the mass of the object in kilograms $$\mathrm{kg}$$, and $$a$$ is the acceleration of the object in meters per second squared$$\frac{\mathrm{m}}{\mathrm{s}^2}$$.

If a box with a mass of 5 kg is pulled with a force of 25 N, what is the subsequent acceleration of the box?

Solution

According to Newtons second law:

$\text{Resultant Force}=\text{mass}\cdot\text{acceleration}$Here, the box is pulled with a force of 25 N, which is our resultant force, and the mass is given to us in the question. If we plug these into the equation for Newtons second law, we can find the magnitude of the acceleration of the box.\begin{align}25\,\mathrm{N}&=5\,\mathrm{kg}\cdot a\\\Rightarrow a&=\frac{25\, \mathrm{N}}{5\,\mathrm{kg}}\\&=5\frac{\mathrm{m}}{\mathrm{s}^2}\end{align}

Newton's first law is a special case of the second law, as clearly if the acceleration of an object is zero then there is resultant force is also zero.

Whilst the definition of Newton's second law given above is more common and is almost always sufficient for solving problems there is a more rigorous and fundamental definition that is useful to know. It states that the resultant force on an object is equal to the time derivative of the object's momentum. Mathematically,$F=\frac{\mathrm{d}p}{\mathrm{d}t}$

Recall that the momentum of an object is $$p=mv$$. This formulation can be used in situations where the mass of the object is non-constant and is more precise as mass is not necessarily a conserved quantity whereas momentum always is.If the mass of an object is a constant then we easily recover the usual equation for Newton's second law\begin{align}F=&\frac{\mathrm{d}p}{\mathrm{d}t}\\=&\frac{\mathrm{d}(mv)}{\mathrm{d}t}\\=&m\frac{\mathrm{d}v}{\mathrm{d}t}\\=&ma\end{align}The final step comes from the definition of acceleration as the time derivative of velocity.

Newton's Third Law

The third and final Newton's law of motion concerns reaction forces. Imagine that you are sitting in a wheelchair, and you push against a wall causing you to roll backward. Why is this? Well when you push against the wall, the wall exerts an equal and opposite reaction force on you causing you to roll backward. This idea is encapsulated by Newton's third law. We call this combination of action and reaction forces, action-reaction pairs.

Newton's Third Law - For every force, there is always a reaction force with the same magnitude acting in the opposite direction.

It is worth noting that the action-reaction pairs do not apply to the same object and are of the same type of force. For example, a book resting on a table experiences a gravitational force from the earth, the reaction force is the gravitational force exerted on the earth by the book.

When a ball bounces on the ground it exerts a downwards force onto the ground, the ground then exerts a force of equal magnitude onto the ball upwards in the opposite direction. This reaction force causes the ball to bounce upwards. This may seem strange as we never see the earth move downwards when we bounce a ball, however, remember that $$F=ma$$, so the huge mass of the earth means that the acceleration of the earth is negligible.

Translational Dynamics Meaning

When a body's position does not change with respect to time, we say it is at rest. However, when a body's position changes with respect to time, we say it is in motion. The study of motion in physics falls into two categories dynamics and kinematics. In kinematics, we are only concerned with things like an object's position, velocity, acceleration, etc. and how it changes over time. Kinematics does not look at the causes of motion and so ignores quantities such as momentum, force, or energy. On the other hand, dynamics looks at the broader picture looking at why an object moves and analyzing motion by looking at the resultant forces on an object or the work done by the object. Newton's Laws of Motion give us an understanding of forces as the cause of motion, from these laws we can then build up a full understanding of the motion of an object. We define two kinds of dynamics, translational dynamics, and rotational dynamics. Rotational dynamics is concerned with objects moving around an axis of rotation such as spinning objects or objects undergoing circular motion.

Fig. 4 - The motion of a Ferris wheel is rotational and falls under rotational dynamics, Wikimedia Commons

Translational dynamics concerns the motion of objects where all parts of the body travel uniformly in the same direction. We can think of it as a kind of sliding where the orientation of the boy does not change whilst the object moves.

For example, a bullet fired from a gun undergoes translational motion as does a block sliding down an inclined plane.

Translational Dynamics Model

Being able to model the motion of an object or group of objects is the central purpose of translational dynamics. In physics, we call an object, or group of objects, under consideration a system. A system can be really simple like a single non-interacting particle, or it can be as complex as a galaxy held together by gravitational interactions.

In translational dynamics, we can simplify things by considering a system's mass to be accumulated entirely within the center of mass, this way forces only act at one point of the system and we don't have to consider how the force acts on each component of the system. We can make this simplification when the properties of the constituent particles are not important to model the behavior of the macroscopic system.

The center of mass of a system is the point of a system where the weighted relative position of the distributed mass of the system is zero. For a system composed of a finite number of point masses, it can be found using the following formula$\vec{r}_{\mathrm{cm}}=\frac{\sum{m_i\vec{r}_i}}{\sum{m_i}}$If the distribution of mass is continuous, like in a single solid object, then we need to integrate instead

$\vec{r}_{cm}=\frac{\int\vec{r}\mathrm{d}m}{\int \mathrm{d}m}$

If a force acts through an object's center of mass it will experience linear translational motion, it's only if the force acts away from the center of mass that rotational motion can occur. This is why objects balance if they're held up at their center of mass.

Fig. 5 - By holding up the toy at its center of mass the toy can balance on the person's finger

In translational dynamics, one of the most commonly occurring problems is calculating the resultant force on an object to understand the direction of an object's acceleration. This involves adding up all the force vectors acting on the object and finding the resultant force vector. Keeping track of all the forces acting on a body is easiest done using a free-body force diagram. These are simple diagrams where each force acting on an object is represented by an arrow pointing in the direction the force acts and accompanied by the magnitude of the force written beside it. Such a diagram is shown below, demonstrating the forces acting on a block sliding down an inclined plane.

Fig.6-Free-body diagram of a block on an incline plane. The friction force acts opposite to the direction of the object's motion, whilst the normal force acts perpendicularly to the surface and the weight acts downwards from the centre of mass.

When multiple forces are acting on an object that are not acting along more than one axis, then we need to select a coordinate system to resolve the forces into their components. By adding the components we can find the resultant force in this coordinate system. In the diagram above, the three forces on the block are not acting in the same direction, so we need to resolve the vectors into their $$x$$ and $$y$$ components to find the resultant force vector.

Notice that the angle between the weight and the normal force is equal to the angle of inclination $$\theta$$. Choosing to resolve the forces into components defined by a coordinate system whose axes are parallel and perpendicular to the surface of the ramp means that we only need to resolve the weight into its components. The friction and normal force are already aligned with the axes of this coordinate system simplifying the calculation. Resolving the weight into these components gives:\begin{align}W_x=& mg\sin(\theta)\\W_y=& mg\cos(\theta)\end{align}

This model of forces and resolved force components set us up to calculate the resultant force acting on the object and its acceleration.

Translation Dynamics Formulas

Once all the forces have been drawn on a free-body diagram and an appropriate coordinate system has been chosen to resolve the force vectors into, we can use Newton's laws to calculate the acceleration of an object.The formula for calculating the resultant force $$F_{\text{net}}$$ can be given as

\begin{align}(F_{\text{net}})_x&=\sum_i (F_i)_x\F_{\text{net}})_y&=\sum_i (F_i)_y\end{align}where \(i is an index for the forces acting on the object.

Returning to the block on the inclined plane, labeling the friction as $$F_{\mu}$$ and the normal force $$F_{\text{norm}}$$ the resultant force is given by

\begin{align}(F_{\text{net}})_x&=F_{\mu}+mg\sin(\theta)\F_{\text{net}})_y&=F_{\text{norm}}+mg\cos(\theta)\end{align} We can use Newton's third law of motion to find the value of the normal force acting on the block. As every action has an equal and opposite reaction force, the normal force must be the reaction force of the plane's surface equal and opposite to the component of the weight acting on the surface. \begin{align}&F_{\text{norm}}=-mg\cos(\theta)\\&\Rightarrow (F_{\text{net}})_y=F_{\text{norm}}+mg\cos(\theta)=0\end{align} This is as we expect as we chose our \(y-axis to be perpendicular to the surface of the inclined plane, and clearly, the block is neither falling through the plane nor levitating above it! If we had instead lined up our $$y$$ axis to be perpendicular to the ground, there would be a force acting downwards.The block is sliding down the plane, so we need to use Newton's second law to find the acceleration of the block along the $$x$$ axis. Recall that$F_{\text{net}}=ma$so, the acceleration along the $$x$$ axis is\begin{align}F_{\mu}+mg\sin(\theta)&=ma_x\\\Rightarrow a_x&=\frac{F_{\mu}}{m}+g\sin(\theta)\end{align}

The value of $$F_{\mu}$$ is determined by the surface's coefficient of friction $$\mu$$and is given by

$F_{\mu}=\mu F_{\text{norm}}=\mu m\cos(\theta)$

This gives a full run-through of how we can use Newton's laws of motion to determine the translational dynamics of an object, so let's apply this method to an explicit example.

Consider a wooden block held up by two cables, as shown in the diagram below, with the wooden block having a mass of $$5.00\mathrm{kg}$$. If the first cable is at an angle of $$\theta_1=-25.0^{\circ}$$ to the vertical, and the second is at an angle of $$\theta_2=45.0^{\circ}$$ what will the tension in each cable be?

Fig.7- By resolving the forces acting on the block we can find the tension in each cable required to hold it up.

First write down all the forces acting on the block. Let the tension in cable one be $$F_1$$ and in cable two $$F_2$$ and obviously the weight will be $$F_{weight}=mg=5.00\cdot9.81=49.0\,\mathrm{N}$$. By choosing our coordinate axes such that the $$y$$ axis is the vertical axis and $$x$$ is the horizontal, we know that, if the block is in equilibrium, the following equations must hold.

\begin{align}(F_1)_x+(F_2)_x&=0\F_1)_y+(F_2)_y&=F_{\mathrm{weight}}=49.0\,\mathrm{N}\end{align} We can express these components using trigonometry\begin{align}F_1\sin(\theta_1)+F_2\sin(\theta_2)&=0\,\mathrm{N}\\F_1\cos(\theta_1)+F_2\cos(\theta_2)&=49.0\,\mathrm{N}\end{align} Re-arranging the first equation and plugging in the angles gives: $F_1=-F_2\frac{\sin(-25^{\circ})}{\sin(45^{\circ})}=0.60F_2$ Substituting this into the second equation \begin{align}0.60&\cdot\cos(-25.0^{\circ})F_2\,\mathrm{N}+F_2\cos(45.0^{\circ})\,\mathrm{N}=49.0\,\mathrm{N}\\\Rightarrow& 0.60\cdot(0.91)F_2\,\mathrm{N}+0.71F_2\,\mathrm{N}=49.0\,\mathrm{N}\\\Rightarrow& F_2=\frac{49}{0.60\cdot0.91+0.71}\,\mathrm{N}=39.0\,\mathrm{N}\end{align}This means the two tensions are \begin{align}F_2&=39.0\,\mathrm{N}\\F_1&=0.60F_2=23.0\,\mathrm{N}\end{align} Translational Dynamics - Key takeaways • In mechanics, an object or group of objects under consideration is called a system • Kinematics looks at how a system moves over time through quantities such as position velocity and acceleration without considering the causes of motion, whereas dynamics considers analyses the causes of motion and is concerned with things like force and energy • When a body moves as a whole and every portion of the body travels the same distance in the same amount of time, then we say the body is in translational motion. On the other hand, if the body is rotating around a fixed axis, this is called rotational motion. • Newton's Laws are the fundamental laws governing the relationship between forces and motion • Newtons first law states that if an object is at rest or in uniform motion unless there is an external force, it will preserve its status. • Newtons second law, force is the product of mass and acceleration. • Newtons third law states that for every action force, there is a reaction force of the same magnitude but in the opposite direction. • We can use free body diagrams to visualize how forces act on a system and to find the resultant force acting on the system's center of mass References 1. Fig. 1 - A box on the ground, StudySmarter Originals 2. Fig. 2 - Person pushing wall, StudySmarter Originals 3. Fig. 3 - Bouncing ball, StudySmarter Originals 4. Fig. 4 - Ferris Wheel (https://commons.wikimedia.org/wiki/File:Ferris_Wheel_-_2590239345.jpg), by seabamirum licensed by CC BY 2.0 (https://creativecommons.org/licenses/by/2.0/deed.en) 5. Fig. 5 - Bird Toy Showing Center of Gravity (https://commons.wikimedia.org/wiki/File:Bird_toy_showing_center_of_gravity.jpg) by APN MJM licensed by CC BY-SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/deed.en) 6. Fig. 6 - Free body force diagram, StudySmarter Originals 7. Fig. 7 - Cable and block, StudySmarter Originals Frequently Asked Questions about Translational Dynamics When a body moves as a whole and every portion of the body travels in the same direction, then we say the body is in translational motion. Examples of translational motion can be a car moving in a straight line, and the path of a bullet out of a gun. When a body moves as a whole and every portion of the body travels in the same direction, then we say the body is in translational motion. On the other hand, if the body is rotating around a fixed axis, this is called rotational motion. Rolling with sliding is an example of both translational and rotational motion. Yes. When a body moves as a whole and every portion of the body travels the same distance in the same amount of time, then we say the body is in translational motion. Final Translational Dynamics Quiz Question Friction is not a contact force. Show answer Answer False. Show question Question A car can brake and stop without the friction force. Show answer Answer False. Show question Question The static friction force is applied while the object is stationary. Show answer Answer True. Show question Question If the force \(\overrightarrow F is exerted on the box pulling it to the right, and its magnitude is increasing with time, the box stays at rest for some time. This is because even if the force is increasing, the force of static friction is increasing as well to cancel the applied force and maintain the box at rest. The force needed to move the box is equal and opposite to the maximum static friction force, $$\mu_{\mathrm s}N$$.

True.

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Question

The type of friction where there is the ___ value of static friction force and the motion is about to start is called the limiting friction.

Maximum.

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The difference between static and kinetic friction force is that the static one is applied while the object is ___, and the kinetic one is applied while the object is ___.

Stationary, moving.

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Fire from a lighter or match is one of the daily life examples of friction.

True.

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Climbers climbing the mountains have nothing to do with the friction.

False.

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There is friction between the ground and the wheels when driving.

True.

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The maximum value of the static friction force is always greater than the magnitude of kinetic friction force.

True.

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For an object at rest, the magnitude of the static friction force is equal to the magnitude of an applied force.

True.

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Question

What is the formula for the frictional force acting on an object as it slides on a surface?

$$F=\mu F_{\text{normal}}$$
where $$\mu$$ is the co-efficient of friction of the surface, and $$F_{\text{normal}}$$ is the normal force of the object on the surface.

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When can we assume that all of a system's mass is concentrated at its center of mass?

When the properties of the system's constituent particles are not important to model the motion of the system.

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What are the two types branches of mechanics that study the motion of systems?

Kinematics and Dynamics.

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If a force is applied at an object's center of mass, it will undergo rotational motion. True or False?

False.

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What is the formula for Newton's second law?

$$F=ma$$

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If a box with a mass of 5 kg is pulled with a force of 20 N, what is the acceleration?

$$4\,\mathrm{m}/\mathrm{s}^2$$.

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If a box is pulled with a force of 35 N and has an acceleration of 7 m/s2, then what is the mass of the box?

5 kg.

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The center of mass for an object with a continuous mass distribution is given as:

$$\vec{r}_{\mathrm{cm}}=\frac{\sum{m_i\vec{r}_i}}{\sum{m_i}}$$.

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Action and reaction forces always act on the same object. True or False?

False.

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The gravitational force acting on a ball as it falls to earth has the same magnitude as the force pulling the earth towards the ball. True or False?

True.

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What does Newton's first law state?

Newton's first law of linear motion states that a body will remain in a state of rest or uniform motion unless acted on by a net external force.

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If a box is pulled with a force of 40 N and has an acceleration of 8 m/s2, then what is the mass of the box?

5 kg

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If a box with a mass of 6 kg is pulled with a force of 24 N, what is the acceleration?

4 m/s2

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The SI unit of force is kg.m/s2  (True/False)

False.

Newton (N)

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What does Newton's second law state?

Force is the product of mass and acceleration.

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A non-inertial frame of reference ____.

has acceleration concerning an inertial reference frame.

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Newton's first law of motion helps to describe the behavior of which property of matter?

Inertia.

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The greater a body’s mass, the less its inertia is. (True/False)

False. The greater a body’s mass, the more the body “resists” being accelerated.

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The net force and acceleration are in the opposite direction. (True/False)

False

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Kate pushes a bottle with a mass of 2 kg to her right along a smooth, level lunch counter. The bottle leaves her hand moving at 2 m/s, then slows down as it slides because of the countertop's constant horizontal friction force. It slides for 1 m before coming to rest. What is the magnitude of the friction force acting on the bottle?

4 N

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Question

Kate pushes a bottle with a mass of 3 kg to her right along a smooth, level lunch counter. The bottle leaves her hand moving at 4 m/s, then slows down as it slides because of the countertop's constant horizontal friction force. It slides for 4 m before coming to rest. What is the magnitude of the friction force acting on the bottle?

6 N

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Question

Kate pushes a bottle to her right along a smooth, level lunch counter. The bottle leaves her hand moving at 5 m/s, then slows down as it slides because of the countertop's constant horizontal friction force. It slides for 5 m before coming to rest. What is the magnitude of the acceleration of the bottle?

-2.5 m/s2

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Question

Kate pushes a bottle to her right along a smooth, level lunch counter. The bottle leaves her hand moving at 4 m/s, then slows down as it slides because of the countertop's constant horizontal friction force. It slides for 2 m before coming to rest. What is the magnitude of the acceleration of the bottle?

-4 m/s2

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Question

If a box with a mass of 8 kg is pulled with a force of 64 N, what is the subsequent acceleration of the box?

8 m/s2

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For every action force, there is a reaction force in the same magnitude acting in the opposite direction. These action-reaction forces always exist in pairs. These forces act on the same object.

False.

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Action and reaction forces have different magnitudes.

False.

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Consider a horse trying to pull a cart. Which of the following is the correct description?

The action-reaction forces are equal but opposite, and they act on the horse such that the net force on the horse is zero, and the horse is unable to move.

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A small car of mass $$300\;\mathrm{kg}$$  is pushing a large truck of mass $$500\;\mathrm{kg}$$  due east on a level road. The car exerts a horizontal force of $$500\;\mathrm{N}$$ on the truck. What is the magnitude of the force that the truck exerts on the car?

$$500\;\mathrm{N}$$.

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Question

Boxes A and B are in contact on a horizontal, frictionless surface. Box A has a mass of $$3\;\mathrm{kg}$$ and box B has a mass of $$2\;\mathrm{kg}$$. A horizontal force of $$30\;\mathrm{N}$$ is exerted on box A. What is the magnitude of the force that box A exerts on box B?

$$30\;\mathrm{N}$$.

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Question

Boxes A and B are in contact on a horizontal, frictionless surface. Box A has a mass of $$5\;\mathrm{kg}$$ and box B has a mass of $$4\;\mathrm{kg}$$. A horizontal force of $$36\;\mathrm{N}$$ is exerted on box A. What is the magnitude of the force that box A exerts on box B?

$$16\;\mathrm{N}$$.

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Which of the following are examples of Newton's Third Law?

A boat's propeller pushes the water away, and the water pushes the boat forward.

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Question

Let's say you are driving a car, and hitting a fly. The car and the fly have action-reaction pairs. Even though they exert the same magnitude of force on each other, would the effects be the same?

If a car exerts $$500\;\mathrm N$$  of force on the fly, the fly exerts the same amount of force on the car as well. However, the fly would die because it is not a force that it can handle.

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Question

Kevin is riding his skateboard and pushes off the ground with his foot. This causes him to accelerate at a rate of (10\;\frac{\mathrm m}{\mathrm s^2}\). He weighs $$600\;\mathrm N$$. How strong was his push off the ground? ($$g=10\;\frac{\mathrm m}{\mathrm s^2}$$)

$$600\;\mathrm N$$.

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Erin is riding her skateboard and pushes off the ground with her foot. This causes her to accelerate at a rate of $$5\;\frac{\mathrm m}{\mathrm s^2}$$. She weighs $$400\;\mathrm N$$. How strong was her push off the ground? ($$g=10\;\frac{\mathrm m}{\mathrm s^2}$$)

$$200\;\mathrm N$$.

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Question

A book exerts a force of $$5\;\mathrm N$$ downward, into a chair that exerts a force of $$10\;\mathrm N$$ downward to the floor it stands on. What is the force that the floor exerts upwards on the chair?

$$15\;\mathrm N$$.

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You stand next to a wall on a frictionless skateboard and push the wall with a force of $$20\;\mathrm N$$. How hard does the wall push on you?

$$20\;\mathrm N$$.

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Question

An archer shoots an arrow. The action force is the bowstring against the arrow. The reaction force is:

Arrow's push against the bowstring.

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A player catches a ball. The action force is the impact of the ball against the player's glove. The reaction force is:

The force the glove exerts on the ball.

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A player hits a ball with a bat. The action force is the impact of the bat against the ball. The reaction force is:

The force of the ball against the bat.

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