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Fundamentals Of Physics
Found in: Page 141

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Short Answer

A loaded penguin sled weighing 80 N rests on a plane inclined at angle θ=200to the horizontal (Fig. 6-23). Between the sled and the plane, the coefficient of static friction is 0.25, and the coefficient of kinetic friction is 0.15. (a) What is the least magnitude of the force parallel to the plane, that will prevent the sled from slipping down the plane? (b) What is the minimum magnitude F that will start the sled moving up the plane? (c) What value of Fis required to move the sled up the plane at constant velocity?

(a) The least magnitude of force F that prevents the sled moving down is 8.6 N

(b) The minimum magnitude of F that will start the sled moving up the plane is 46 N

(c) The value of F, that is required to move the sled up the plane at constant velocity is 39 N

See the step by step solution

Step by Step Solution

Step 1: Given

Weight, W=80 N

Coefficient of static friction, μs=0.25

Coefficient of kinetic friction, μk=0.15

Inclined angle, θ =200

Step 2: Determining the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Use the Newton's 2nd law of motion along vertical and horizontal direction.

Formula:

Fnet=ma

where, F is the net force, m is mass and a is an acceleration.

Step 3: Determining the free body diagram

Free body Diagram of sled over the inclined plane:

Step 4: (a) Determining the least magnitude of the force parallel to the plane

First, from the weight, calculate the mass of the system,

W=mgm=Wg =80 N9.81 =8.2kg

By using Newton’s 2nd law along vertical direction (along y),

Fy=may

since, block is not moving upward, ay=0

N-Fgcosθ=0N=Fgcosθ

Relation between static frictional force and normal force is,

fs=μsN=NsFgcosθ (i)

When sled is moving down frictional force is upward,

Now, applying Newton’s 2nd along the x direction, for fig. a,

Fs=max

since, block is not moving, localid="1654149316961" ax=0

F+fs-Fgsinθ=0F=fgsinθ-fs

By using equation (i) in above equation,

F=Fgsinθ-μsFgcosθF=Fg(sinθ-μscosθ) =mg(sin20-(0.25)cos20) =8.6 N

Thus, the least magnitude of force F that prevents the sled moving down is 8.6 N

Step 5: (b) Determining the minimum magnitude of F that will start the sled moving up the plane

To find the F to start the sled moving up is, again we have to use Newton’s 2nd law along y and x direction, but in this case as the sled moving up frictional force is downward.So, refer to fig. b.

Along vertical direction (y),

Fy=may

since block is not moving upward,ay=0

N-Fgcosθ=0N=Fgcosθ

Then static frictional force,

fs=μsN=μsFgcosθ

Along horizontal direction (x),

Fx=max

since, block is not moving, ax=0

F-fs-Fgsinθ=0F=Fgsinθ+fs =Fg(sinθ+μscosθ) =46 N

Hence, the minimum magnitude of F that will start the sled moving up the plane is 46 N.

Step 6: (c) Determining thevalue of F, that is required to move the sled up the plane at constant velocity

To find F to move the sled up the plane at constant velocity, use kinetic frictional force,

fk=μkN =μkFgcosθ

Along horizontal direction (x),

Fx=max

since, block is moving with constant velocity, ax=0

F-fk-Fgsinθ=0F=Fgsinθ+fk =Fg(sinθ+μkcosθ) =39 N

Hence, the value of F, that is required to move the sled up the plane at constant velocity is 39 N

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