Stupid Stubborn Mule and his Kinetic/Static Friction

[Phoenixtears]

SOLVED

Homework Statement

A stubborn, 120 kg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around the mule and pulls with his maximum force of 770 N. The coefficients of friction between the mule and the ground are µs = 0.8 and µk = 0.5. Is the farmer able to move the mule?

no



If so, how much force in excess of the required amount did the farmer pull? If not, how much more force does the farmer require?
N

Homework Equations

F= Ma
F(fric)= (Coefficient of friction)*N(normal force)

The Attempt at a Solution

I got the answer to the first part of this question through guessing, so I'm not sure if there is actually work invovled there that needs to be used for part 2 that I therefore don't have due to the fact that I guessed. Anyway, I'm not sure how to approach this problem. Using those equations I got answers separately for static friction and kinetic friction, but I don't think either of those is what I'm looking for. My thought is that I can just disregard kinetic for a second because using 770N, only static friction is involved.

Is there a different way to approach this problem??

Thank you so much in advance!

~Phoenix

PS- I do realize that my net force is zero, for in my force diagram all the vectors should equal each other. But, then how is one to get the amount of force needed to actually pull the mule?

 

[tiny-tim]

I got the answer to the first part of this question through guessing, so I'm not sure if there is actually work invovled there that needs to be used for part 2 that I therefore don't have due to the fact that I guessed. Anyway, I'm not sure how to approach this problem. Using those equations I got answers separately for static friction and kinetic friction, but I don't think either of those is what I'm looking for. My thought is that I can just disregard kinetic for a second because using 770N, only static friction is involved.

Hi Phoenix!

Yes, you're right … the kinetic friction is irrelevant.

I do realize that my net force is zero, for in my force diagram all the vectors should equal each other. But, then how is one to get the amount of force needed to actually pull the mule?

In exam questions like this, you should always assume that the acceleration is negligible … the mule moves, but only just … in other words, the acceleration is zero.

Show us what you think the answer is.

 

[Phoenixtears]

Hi Phoenix!

In exam questions like this, you should always assume that the acceleration is negligible … the mule moves, but only just … in other words, the acceleration is zero.

The only relevant equation that involes the acceleration is F=ma. However, showing that the acceleration is zero wouldn't aid us in that equation. Hmmm...

All right, a different view...

Using the fact that acceleration is so low would mean that the total force comes out to be zero. Therefore the answer would be zero. But that doesn't seem right at all. 770N could be 1000N away from actually being able to move the mule just an inch.

µs = 0.8
Using this seems like a good idea. But when put into the second equation, the maximum static friction comes out to be 96. (.8*120). I have no idea where to go.
Using this

 

[Doc Al]

But when put into the second equation, the maximum static friction comes out to be 96. (.8*120).

What's the maximum static friction force that the mule can generate? (Correct your equation above. What's the normal force? 120 kg is the mule's mass.)

Compare this to the applied force of 770N and then decide whether he can move the mule or not.

 

[Phoenixtears]

GAR! Darn it! I don't know how many times I've substituted mass for weight.

In other words, all I have to do is multiply 120*9.8 to get the weight. (Which is 1176) Then multiply that by .8, to give me the maximum static friction. 941-770=171.

Thank you so much!

Oh, a friend asked about what would happen if the man could move the mule and we were trying to find how much extra force was exerted. Wouldn't that mean taking the WEIGHT (got it this time) being 1176, then multiplying by .5 (the kinetic energy), equaling 588. 770-588=182. Would that be correct?

 

[Doc Al]

Oh, a friend asked about what would happen if the man could move the mule and we were trying to find how much extra force was exerted. Wouldn't that mean taking the WEIGHT (got it this time) being 1176, then multiplying by .5 (the kinetic energy), equaling 588. 770-588=182. Would that be correct?

Exactly. (But that's the coefficient of kinetic friction, not kinetic energy. )

 

[Phoenixtears]

Right, right, of course. :)

Thank you so much for all of the help.

~Phoenix