# A stubborn, 100.0kg mule sits down and refuses to move. To drag the mu

1. Apr 17, 2013

### cmkc109

1. The problem statement, all variables and given/known data

A stubborn, 100.0kg 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 826.0N. The coefficients of friction between the mule and the ground are s=0.800 and k=0.220. What is the net force on the mule?

3. The attempt at a solution

ok, so the original question was 144kg mule, but i changed the numbers around to make the force greater than static friction.

100x 9.8 x 0.8 =784N

since the applied force is greater than static force, we know that mule is moving.
BUT, the net force is not 826 - 784 right because we need to use kinetic friction.

so 100 x 9.8 x 0.22 =215N. so net force is 826 - 215 =611N

BUT if the question what is the net force when the mule just started to move,
will it be 826 - 784 = 42N?

but that's such a big difference of net force..

Can someone check both of my calculations and reasoning?

2. Apr 18, 2013

### Simon Bridge

784N is what was needed to "unstick" the mule from the ground.

If the farmer takes some time to build up to max force, then the mule unsticks at 784N, then, right away, starts to accelerate under an unbalanced (784-215)N. The acceleration won't be constant because the farmer is not yet at full pulling force.

But what the question writer would want to see is probably what you did.
"just started to <thingy>" usually means the bit just before the <thingy>.

3. Apr 18, 2013

### cmkc109

"accelerate under an unbalanced (784-215)N."
do you mean 826 - 215?

4. Apr 18, 2013

### Simon Bridge

No ...
... i.e. when the mule unsticks, the farmer is not pulling with max force yet.

If you were the farmer you'd start out by, first, taking up the slack, then taking strain, and then you'd pull harder and harder until the mule starts moving, wouldn't you?