# Equilibrium force

1. Mar 18, 2005

### UrbanXrisis

so first I found the angle...
$$tan\theta=\frac{O}{A}=\frac{0.5}{6}$$
$$\theta=4.7636416$$

I then needed to find the force of the hypotnuse...
$$sin\theta=\frac{O}{H}$$
$$H=\frac{250N}{sin4.7636416}$$

$$H=3010.3987N$$

is this correct?

Last edited by a moderator: May 1, 2017
2. Mar 18, 2005

### witze

I didn't bother going through your work, but I got the same answer, 3010.4 N. So I assume it's correct. Also, the distances in your problem were given to 2 significant digits, so realistically the answer is correct to only 2 sig dig.

Last edited: Mar 18, 2005
3. Mar 18, 2005

### physmurf

That looks good to me. Assuming all of your calculations are correct, that seems to be the correct method for solving this problem.

4. Mar 19, 2005

### UrbanXrisis

it just seems amazing that if you applied only 500 newtons, you will get a resultant force of over 6 times of what you applied.

how can that be?

5. Mar 19, 2005

### whozum

Mechanical advantage. You are applying force against a much larger angle than the force applied against the car. The difference in this angle will result in the force on the car being much greater.