# Equalibrium problem ( i think)

## Homework Statement

Q) To pull a car from a ditch, the driver ties one end of a rope to the car and the other end to a tree 12.5 m away and then pulls sideways at the midpoint of the rope with a 475 N force. How much force is exerted on the car when the driver has pulled the rope 75.0 cm to one side ?

## The Attempt at a Solution

My solution assuming this is a torque issue.
Assuming> the ends are 12.5 meters apart.one end tied to the car, the other tied to the tree.
> driver is pulls the rope 75 cm from the middle 6.25meters.

So the radius = 12.5metres/2 = 6.25meters
Find Ɵ
0.12 = sin Ɵ.
Ɵ = 6.89
The torque is rFsinƟ = 6.25m ( 475N) sin6.89 = 356Nm.

My solution assuming this is an equilibrium issue:

Assuming> the ropes ends are 12.5 meters apart.
> The ropes directions are one end is to the east and the other is to the west
> assuming the driver is pulling the car 75 cm west of north.
So the radius = 12.5metres/2 = 6.25meters
Find
.75metes / 6.25 meters = 0.12 = cos Ɵ.
Ɵ = 83.1
westward - The driver force on the car will be 475Ncos83.1= 57N
Northward - The driver force on the car will be 475Nsin83.1= 471N

Last edited:

Related Introductory Physics Homework Help News on Phys.org
Why are you using the term 'torque'? And do you assume the vehicle has moved a little when the puller has moved the center of the rope 75cm? Or has the rope stretched?

I like Serena
Homework Helper
Welcome to PF, charlene w! I'm afraid torque is not applicable in this problem.

I recommend you draw a picture of the rope.
Draw vectors on it representing the 3 forces on the rope.
In particular the car and the tree have a reactive force on the rope.

For equilibrium the forces must cancel each other.
From that you can calculate the force on the car.

Thanks for the replies! I will give your suggestion some thought.