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Tension on a string attached to two objects

  • #1
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Homework Statement


A person would like to pull a car out of a ditch. This person ties one end of a chain to the car's bumper and wraps the other end around a tree so that the chain is taut. The person then pulls on the chain perpendicular to its length. If the distance between the car and tree is 5.0 m and the length of the chain between the car and the tree is 5.2 m and the person can pull with 100LB of force, what force can the chain exert on the car?

Homework Equations


Basic trigonometry

The Attempt at a Solution



Let d_1, d_2 be 5.0 m and 5.2 m respectively. Let F_1 be 100 LB.
Draw a pair of congruent triangles by connecting a line from the person to the ground. This gives two right triangles with base (1/2)d_1 and hypotenuse (1/2)d_2. You can calculate the angle between the base and hypotenuse to be about 15.94 degrees which we may call theta. Then F_T = csc(theta) * F_1 = 364 N. The book gets about half of that at 180 N.
 

Answers and Replies

  • #2
haruspex
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Homework Statement


A person would like to pull a car out of a ditch. This person ties one end of a chain to the car's bumper and wraps the other end around a tree so that the chain is taut. The person then pulls on the chain perpendicular to its length. If the distance between the car and tree is 5.0 m and the length of the chain between the car and the tree is 5.2 m and the person can pull with 100LB of force, what force can the chain exert on the car?

Homework Equations


Basic trigonometry

The Attempt at a Solution



Let d_1, d_2 be 5.0 m and 5.2 m respectively. Let F_1 be 100 LB.
Draw a pair of congruent triangles by connecting a line from the person to the ground. This gives two right triangles with base (1/2)d_1 and hypotenuse (1/2)d_2. You can calculate the angle between the base and hypotenuse to be about 15.94 degrees which we may call theta. Then F_T = csc(theta) * F_1 = 364 N. The book gets about half of that at 180 N.
Consider the point on the chain where the person is pulling. Draw an FBD for that. How many forces are there?
 
  • #3
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Consider the point on the chain where the person is pulling. Draw an FBD for that. How many forces are there?
Of course I've done that. If we ignore the frictional forces with the ground as the book usually does in a situation like this, and cancel the gravitational and normal forces, there are two forces acting on the person along the two chains.
 
  • #4
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Of course I've done that. If we ignore the frictional forces with the ground as the book usually does in a situation like this, and cancel the gravitational and normal forces, there are two forces acting on the person along the two chains.
Consider the point on the chain where the person is pulling. Draw an FBD for that. How many forces are there?
On second thought, I don't think I should ignore the frictional force with the ground. The frictional force with the ground is acting say downward, and for static equilibrium, the upward forces from the chains have to cancel, and I think I get the right answer this way. Is that how you thought about it?
 
  • #5
haruspex
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Of course I've done that. If we ignore the frictional forces with the ground as the book usually does in a situation like this, and cancel the gravitational and normal forces, there are two forces acting on the person along the two chains.
Ok, but I would prefer to view it as those two forces, each FT, plus the pull F1 from the person all acting at a point on the chain. So that gives three forces altogether. Consider the force balance in the direction in which the person pulls.
 
  • #6
haruspex
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On second thought, I don't think I should ignore the frictional force with the ground. The frictional force with the ground is acting say downward, and for static equilibrium, the upward forces from the chains have to cancel, and I think I get the right answer this way. Is that how you thought about it?
There's no indication that the chain remains in contact with the ground. It would be unlikely. Anyway, there would be no normal force from the ground, so no friction.
 
  • #7
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OK, I see your point. The friction argument shouldn't matter anyways, because the result still holds in a friction free environment. Of course what I meant was the net forces acting on the person, so friction from the ground plus the two Tensions all sum to 0. But, I like the way you are reasoning here.
 

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