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

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Homework Help Overview

The problem involves a scenario where a farmer attempts to drag a 100.0 kg mule that is initially stationary due to static friction. The discussion centers around calculating the net force acting on the mule when the farmer applies a pulling force of 826.0 N, considering both static and kinetic friction coefficients.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations for static and kinetic friction forces, questioning the net force when the mule begins to move. There is also exploration of the implications of the farmer's pulling force and the transition from static to kinetic friction.

Discussion Status

Participants are actively engaging with the calculations and reasoning presented. There is a focus on clarifying the conditions under which the net force is calculated, particularly regarding the transition from static to kinetic friction. Some participants are questioning the assumptions made about the farmer's pulling force and the timing of the mule's movement.

Contextual Notes

There is a noted difference in calculations based on whether the net force is considered at the moment the mule starts to move or after it has already begun moving. The discussion reflects on the dynamics of force application and frictional forces in this context.

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Homework Statement



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?

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?
 
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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>.
 
Simon Bridge said:
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>.


"accelerate under an unbalanced (784-215)N."
do you mean 826 - 215?
 
cmkc109 said:
"accelerate under an unbalanced (784-215)N."
do you mean 826 - 215?
No ...
me said:
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.
... 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?
 

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