# Pulling a box up a ramp

1. Sep 19, 2014

### Statik

1. The problem statement, all variables and given/known data

A girl of mass mg = 50 kg is pulling a sled up a slippery slope. The coefficient of friction between the girl's boots and the slope is µs = 0.155; the friction between the sled and the slope is negligible. The girl can pull the sled up the slope with a ≤ amax = 0.030 m/s2 before she begins to slip. Assume the rope connecting the girl to the sled is kept parallel to the slope at all times. The angle of the slope is θ = 8°.

2. Relevant equations

3. The attempt at a solution

I have already worked out the maximum value of the force of friction on the girl as:
50kg *9.8 m/s^2 = 490N
Mg(j-hat) = COS 8 * 490N = 485.23N
Fs = Us * Mg(jhat) = 0.155 * 485.23N = 75.21N
Gravity Force on Girl, down the ramp = Sin 8 * 490N = 68.19N

Net force unused by girl is75.21N - 68.19N = 7.02N
Net acceleration is 0.03m/s^2 so, using remaining force, 7.02N / 0.03m/s^2 = 234kg
Total box should be 234kg / sin 8 = 1681 kg but that isn't right. I know my total friction for the girl and I know it's right . . . what am I missing in my calculation of the box weight?

Thanks!
Rod

2. Sep 20, 2014

### PhanthomJay

You did well I in determining the tension in the rope of 7 N, but then you did not analyze the net force acting on the sled correctly. The tension acts up the slope on the sled, but the component of its unknown weight acts down the slope. Try your F_net = ma equation again with a good free body diagram. Like you did when analyzing the forces on the girl.