Pulling a box up a ramp

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