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

  • Thread starter Thread starter cmkc109
  • Start date Start date
  • Tags Tags
    Drag
AI Thread Summary
The discussion revolves around calculating the net force acting on a 100.0kg mule being pulled by a farmer with a maximum force of 826.0N. The static friction force required to move the mule is calculated to be 784N, indicating that the applied force exceeds this threshold, allowing the mule to start moving. Once in motion, the kinetic friction force is determined to be 215N, leading to a net force of 611N when the mule is moving. There is some confusion regarding the net force calculation at the moment the mule begins to move, with a suggestion that it could be 42N if only considering static friction. The conversation emphasizes the importance of understanding the transition from static to kinetic friction in this scenario.
cmkc109
Messages
103
Reaction score
0

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?
 
Physics news on Phys.org
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?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
3
Views
13K
Back
Top