• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q89P

Expert-verified
Fundamentals Of Physics
Found in: Page 147

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

A filing cabinet weighing 556 N rests on the floor. The coefficient of static friction between it and the floor is 0.68 , and the coefficient of kinetic friction is 0.56 . In four different attempts to move it, it is pushed with horizontal forces of magnitudes (a) 222 N , (b) 334 N , (c) 445 N , and (d) 556 N . For each attempt, calculate the magnitude of the frictional force on it from the floor. (The cabinet is initially at rest.) (e) In which of the attempts does the cabinet move?

(a) The magnitude of the frictional force is f = 222 N .

(b) The magnitude of the frictional force is f = 334 N .

(c) The magnitude of the frictional force is f = 311 N .

(d) The magnitude of the frictional force is f = 311 N .

See the step by step solution

Step by Step Solution

Step 1: Given data:

Weight of the cabinet, W = 556 N

Coefficient of static friction between the cabinet and the floor, μs=0.68

Coefficient of kinetic friction, μk=0.56

Step 2: Understanding the concept:

In order to move a filing cabinet, the force applied must be able to overcome

the frictional force.

Apply Newton’s second law

Fpush-f=Fnet =ma

If you find the applied force Fpush to be less than fs,max , the maximum static frictional force, our conclusion would then be “no, the cabinet does not move” (which means is actually zero and the frictional force is simply f=Fpush ). On the other hand, if you obtain a > 0 then the cabinet moves (so f=fk ).

For fs,max and fk use Eq. 6-1 and Eq. 6-2 (respectively), and in those formulas set the magnitude of the normal force to the weight of the cabinet.

Step 3: Calculate the maximum static friction and kinetic friction:

The maximum static frictional force is,

fs,max=μsFN

Here, the weight of the cabinet will be balanced by the normal force. Therefore,

fs,max=μsW

And the kinetic frictional force is,

fk=μkfk =μkW =0.56556N =311N

Step 4: (a) Define the magnitude of the frictional force with horizontal force of magnitude 222 N :

Calculate the magnitude of the frictional force on the cabinet from the floor when it is pushed with horizontal force of magnitudes, which is,

.Fpush=222 N

Here, the magnitude of the horizontal pushing force is less than the maximum static force. Therefore,

Fpush<fs,max222 N<378 N

Hence, the cabinet does not move. So, acceleration of the cabinet will be zero.

a=0 m/s2

The frictional force is,

Fpush-f=ma =m0 m /s2 =0f=Fpush =222 N

Here, the magnitude of the frictional force between the carbonate and the floor is f = 222N .

Step 5: (b) Determine the magnitude of the frictional force with horizontal force of magnitude 334 N :

Calculate the magnitude of the frictional force on the cabinet from the floor when it is pushed with horizontal force of magnitudes, which is

Fpush=334 N

Here, the magnitude of the horizontal pushing force is less than the maximum static force. Thus,

Fpush<fs,max334 N<378 N

Hence, the cabinet does not move. So, acceleration of the cabinet will be zero.

a=0 m/s2

The frictional force is,

Fpush-f=ma =0f=Fpush =334 N

Step 6: (c) Define the magnitude of the frictional force with horizontal force 445 N :

Calculate the magnitude of the frictional force on the cabinet from the floor when it is pushed with horizontal force of magnitudes, which is,

Fpush=445N.

Here, the magnitude of the horizontal pushing force is less than the maximum static force. Thus,

Fpush>fs,max445 N>378 N

Hence, the cabinet will move. So, the frictional force will be he kinetic friction. Hence, the friction force in this case between the cabinet and the floor is,

f=Fk =311 N

Step 7: (d) The magnitude of the frictional force with horizontal force 556 N :

Calculate the magnitude of the frictional force on it from the floor when it is pushed with horizontal force of magnitudes, which is,

Fpush=556 N.

Again, you have

Fpush>fs,max556 N >378 N

Which means the cabinet moves.

Hence, the cabinet will move. So, the frictional force will be he kinetic friction. Hence, the friction force in this case between the cabinet and the floor is,

f=fk =311 N

Step 8: (e) Find out in which of the attempts does the cabinet move:

As in part (c) and (d) you have Fpush>fs,max which means the cabinet moves.

Hence, the cabinet moves in (c) and (d).

Most popular questions for Physics Textbooks

Icon

Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.