Calculating Applied Force for a Moving Block on a Table

[Alyona]

Homework Statement

You gradually apply a force to a 10. kg wood block on a wood table. What will the applied force on the block be once it is moving? The coefficients of static and kinetic friction are 0.5 and 0.3, respectively. The acceleration due to gravity is 9.8 m/s2.

Homework Equations

N = w= mg
Fk = μkN
Fsmax = μsN

The Attempt at a Solution

N = w = mg = (10 kg)(9.8 m/s2) = 98 N
Fsmax = μsN = (0.5)(98 N) = 49 N (wrong!)

Is this correct?
N = w = mg = (10 kg)(9.8 m/s2) = 98 N
Fk = μkN = (0.3)(98 N) = 30 N

 

[drvrm]

N = w = mg = (10 kg)(9.8 m/s2) = 98 N
Fsmax = μsN = (0.5)(98 N) = 49 N (wrong!)

Is this correct?
N = w = mg = (10 kg)(9.8 m/s2) = 98 N
Fk = μkN = (0.3)(98 N) = 30 N

the frictional force calculated seems to be correct.
now, as your external pulling force gets to limiting friction condition , the body will start to move. as one is still pulling with 49 N
and the motion has started. the question is what will happen now?

 

[haruspex]

the question is what will happen now?

Not as I read it. It asks for the applied force, not the net force.
The applied force the instant before it moved was 49N. We are given no reason to suppose it has been reduced.
In practice, it rather depends how the force was applied. If pushed by hand or attached by a string over a pulley to a weight then the resulting acceleration of the external mass will reduce the applied force. But if applied by increasing the tension in an elastic string then, initially at least, the force will not change.

 

[CWatters]

+1

The block doesn't move until the applied force overcomes the max static friction which is 49N. After that the _friction force_ reduces to 30N but the _applied force_ could stay the same, in which case the block accelerates.

If the book answer is 30N then it's a badly worded question.

 

[Alyona]

+1

The block doesn't move until the applied force overcomes the max static friction which is 49N. After that the _friction force_ reduces to 30N but the _applied force_ could stay the same, in which case the block accelerates.

If the book answer is 30N then it's a badly worded question.

I think I get it now. First the wood block has to overcome the static friction force to start moving and then it has to overcome the kinetic friction force to keep moving.

 

[CWatters]

Yes.

Once moving if the applied force is greater than the kinetic friction the net force is >0 so it acelerates, if lower the net force is <O (negative/opposite direction) so it decelerates (to a stop if nothing else changes).

Its quite common for the coefficient of static friction to be higher than the coefficient of kinetic friction... so you push on something harder and harder until suddenly it moves, friction reduces and you fall on your face because you can't reduce the force you are applying fast enough.

 

[haruspex]

I think I get it now. First the wood block has to overcome the static friction force to start moving and then it has to overcome the kinetic friction force to keep moving.

Yes, but the question as stated in post #1 does not ask what force is necessary to keep it moving. It asks what the applied force will be when it is moving.

The short answer is that it can be whatever you like: we are only told what force is applied up to the moment it moves; we are not told what force is applied subsequently.
The more reasonable interpretation is to assume that the applied force continues, as before, only to change gradually. In this case your answer of 49N is correct.

So, does post #1 state the question exactly, word for word?

 

[Alyona]

You gradually apply a force to a 10. kg wood block on a wood table. What will the applied force on the block be once it is moving? The coefficients of static and kinetic friction are 0.5 and 0.3 respectively. The acceleration due to gravity is 9.8 m/s2.

Selected Answer:
d. 49. N

Answers:
a. 3.0 N
b. 5.0 N
c. 30. N
d. 49. N

Response Feedback: Incorrect. Remember, the friction force is related to the normal force.

 

[CWatters]

Response Feedback: Incorrect. Remember, the friction force is related to the normal force.

The Normal Force is constant in this problem = 10kh * 9.81m/s/s = 98.1N.

The friction force is...

When stationary it increases from zero to a maximum of 98.1 * 0.5 = 49N
Once moving it reduces to 98.1 * 0.3 = 30N

BUT as we have pointed out the problem doesn't ask about the friction force it asks about the applied force. Just before it moves the applied force will be 49N. Just after it starts moving the applied force could be anything.

Show this thread to your teacher. The problem statement should be rewritten something like this...

"You gradually apply an increasing force to a 10. kg wood block on a wood table. The coefficients of static and kinetic friction are 0.5 and 0.3 respectively. The acceleration due to gravity is 9.8 m/s2.

Q1) What will the applied force on the block be just before the block starts moving? A: 49N
Q2) Once moving, what applied force is required to keep the block moving at a constant velocity? A: 29N"

 

[Alyona]

Yes, I agree it needs to be reworded. Thanks for your help!