# dynamics question

• Oct 3rd 2006, 03:07 AM
scorpion007
dynamics question
hi, is this question wrong or am i wrong?
A crate of mass 100 kg rests on a rough horizontal factory floor. The coefficient of friction between the floor and the crate is 1/7. A storeman applies a horizontal force of F newtons to the crate in order to move it to a different location.
Gravity is 9.8 ms^-2

a) Show that if F = 120, the crate will not move.

Now, for the crate not to move, doesnt F == Friction ? If so, i got Fr = 140, not 120.
F = mu * R = 1/7* 100g = 140

So whats wrong?
• Oct 3rd 2006, 04:14 AM
topsquark
Quote:

Originally Posted by scorpion007
hi, is this question wrong or am i wrong?
A crate of mass 100 kg rests on a rough horizontal factory floor. The coefficient of friction between the floor and the crate is 1/7. A storeman applies a horizontal force of F newtons to the crate in order to move it to a different location.
Gravity is 9.8 ms^-2

a) Show that if F = 120, the crate will not move.

Now, for the crate not to move, doesnt F == Friction ? If so, i got Fr = 140, not 120.
F = mu * R = 1/7* 100g = 140

So whats wrong?

Pet peeve warning!
Quote:

Originally Posted by scorpion007
Acceleration due to Gravity is 9.8 ms^-2

Ah. Better now! Pet peeve over. :D

You did it right, but your interpretation was wrong. (Note the section in red letters below.) Let me explain in detail:

Okay, my sketch of the Free-Body Diagram has a weight w acting downward, a normal force N acting upward, an applied force F acting horizontally to the right on the box, and a static friction force f acting to the left. I have a +x direction to the right and a +y direction upward.

Applying Newton's 2nd in the y direction we get:
(Sum)Fy = N - w = ma_y = 0 (since the box is not accelerating upward.)
Thus N = w = mg

Applying Newton's 2nd in the x direction we get:
(Sum)Fx = F - f = ma_x = 0 (since the box is not accelerating horizontally.)
I'm going to calculate the maximum force we can apply to the box such that the box will not move. In this case the static friction force will be equal to (mu)N. (It is generally less than or equal to this amount.) This should be greater than 120 N if the box is to be stationary.

Thus F = f = (mu)N = (mu)mg = (1/7)*100*9.8 = 140 N.

Since we can apply up to 140 N before the box moves and we are only applying 120 N, the box will be stationary.

-Dan
• Oct 3rd 2006, 06:32 PM
scorpion007
ahh i see, so instead of me saying F == Friction for it not to move i should have said:
F <= Friction, correct?
Sorry about the gravity thing, i was rushing it a bit, i do understand its acceleration though.
• Oct 4th 2006, 04:37 AM
topsquark
Quote:

Originally Posted by scorpion007
ahh i see, so instead of me saying F == Friction for it not to move i should have said:
F <= Friction, correct?
Sorry about the gravity thing, i was rushing it a bit, i do understand its acceleration though.

Don't worry about the gravity thing, I've heard professors say it that way, too. :eek:

Yes, the whole problem was that you were trying to show that the applied force was less than the maximum possible static friction. The other way to look at the problem would be to calculate how much static friction was keeping the box in place, then showing that this is less than the maximum. The two approaches involve about the same amount of work.

-Dan
• Oct 4th 2006, 04:42 AM
CaptainBlack
Quote:

Originally Posted by topsquark
Don't worry about the gravity thing, I've heard professors say it that way, too. :eek:

Yes, the whole problem was that you were trying to show that the applied force was less than the maximum possible static friction. The other way to look at the problem would be to calculate how much static friction was keeping the box in place, then showing that this is less than the maximum. The two approaches involve about the same amount of work.

If the crate does not move there is no work involved:cool:

RonL
• Oct 4th 2006, 04:45 AM
topsquark
Quote:

Originally Posted by CaptainBlack
If the crate does not move there is no work involved:cool:

RonL

Whatever is causing the force to try to move the crate is doubtless converting one form of energy to another to do so, so there is an amount of work being supplied to the system. It simply has no effect except to increase the entropy of the surrounding space. :D

-Dan
• Oct 4th 2006, 05:11 AM
CaptainBlack
Quote:

Originally Posted by topsquark
Whatever is causing the force to try to move the crate is doubtless converting one form of energy to another to do so, so there is an amount of work being supplied to the system. It simply has no effect except to increase the entropy of the surrounding space. :D

Mmmmm.. entropy

RonL