# critical thinking vectors

• Jan 6th 2007, 06:38 PM
sinewave85
critical thinking vectors
Jeff is trying to use a long piece of rope to remove a car from some deep mud. Which method should he use to exert more force on the car? Assume that Jeff will exert the same amout of force using either method. Explain your answer.
a. Tie one end of the rope to the car and pull.
b. Tie one end of the rope to the car and the other end to a nearby tree. Then push on the rope perpendicular to it at a point about halfway between the car and the tree.

I think it must be "b", but I don't know exactly how to explain why.
Ok, edit. I did some calculations: apply a force of 10N perpendicular to the rope, displacing it 1 degree, and the tension in the rope on either side of the displacement is about 573N. The tension goes down with each degree of displacement. I dont know, however, if I am sort of missing the point. I can prove that it works, but why? This is a math class, not physics -- resolution of vectors. Looking at my drawings, the applied force is the leg of a right triangle, and the tension is the hypotenuse. The other leg does not really seem to enter into it -- dissipated force? Anyway, for the purpose of "Explain your answer" -- the hyp is always longer than a leg, thus the benefit?
• Jan 6th 2007, 09:02 PM
ticbol
Quote:

Originally Posted by sinewave85
Jeff is trying to use a long piece of rope to remove a car from some deep mud. Which method should he use to exert more force on the car? Assume that Jeff will exert the same amout of force using either method. Explain your answer.
a. Tie one end of the rope to the car and pull.
b. Tie one end of the rope to the car and the other end to a nearby tree. Then push on the rope perpendicular to it at a point about halfway between the car and the tree.

I think it must be "b", but I don't know exactly how to explain why.
Ok, edit. I did some calculations: apply a force of 10N perpendicular to the rope, displacing it 1 degree, and the tension in the rope on either side of the displacement is about 573N. The tension goes down with each degree of displacement. I dont know, however, if I am sort of missing the point. I can prove that it works, but why? This is a math class, not physics -- resolution of vectors. Looking at my drawings, the applied force is the leg of a right triangle, and the tension is the hypotenuse. The other leg does not really seem to enter into it -- dissipated force? Anyway, for the purpose of "Explain your answer" -- the hyp is always longer than a leg, thus the benefit?

Yes, method B will exert more pulling force on the car.

Let's say Jeff can pull a maximum of F newtons force pulling or pushing.

In method A, the maxinum pulling force on the rope will be F newtons.

In method B, the pulling force on the rope is much more than F newtons.
The tension on the rope is the pulling force on the rope, say T newtons. The F pushing the rope at about the rope's midpoint will impart tension T on the rope.
Get at the intersection of the F and the two opposing T's where the rope is being pushed. One T points to the tree; the other T points to the car.

The components (we call Tp) of these two T's that are parallel to the F will equal or neutralize the F. So,
2(Tp) = F
Tp = F/2 = (0.5)F newtons.

The components (we call Tr) of these two opposing T's perpendicular to F will neutralize each other. [You thought of them as "dissipated forces".]

Let us say the rope is bent 178 degrees at the point of pushing. Then a T is 89 degrees from the direction of F. Hence,
T*cos(89deg) = 0.5 F
T = (0.5 F) / cos(89deg)
T = (28.65)F newtons -----which is a lot more than F newtons.
• Jan 7th 2007, 12:24 PM
sinewave85
Quote:

Originally Posted by ticbol
Yes, method B will exert more pulling force on the car.

Let's say Jeff can pull a maximum of F newtons force pulling or pushing.

In method A, the maxinum pulling force on the rope will be F newtons.

In method B, the pulling force on the rope is much more than F newtons.
The tension on the rope is the pulling force on the rope, say T newtons. The F pushing the rope at about the rope's midpoint will impart tension T on the rope.
Get at the intersection of the F and the two opposing T's where the rope is being pushed. One T points to the tree; the other T points to the car.

The components (we call Tp) of these two T's that are parallel to the F will equal or neutralize the F. So,
2(Tp) = F
Tp = F/2 = (0.5)F newtons.

The components (we call Tr) of these two opposing T's perpendicular to F will neutralize each other. [You thought of them as "dissipated forces".]

Let us say the rope is bent 178 degrees at the point of pushing. Then a T is 89 degrees from the direction of F. Hence,
T*cos(89deg) = 0.5 F
T = (0.5 F) / cos(89deg)
T = (28.65)F newtons -----which is a lot more than F newtons.

Ok, I see! It works because the exerted force only has to counteract the Tp force -- and when the angle on the rope is very obtuse, the Tp component is very small. I had not "found" the Tp force in my drawings, only T, Tr, and F, and I could tell that something was missing, but not what. Stupid me. Anyway, thanks for the help.