How Much Can the Sled Weigh If a Girl Pulls It Up a Slope Without Slipping?

[Kingsley]

Homework Statement

A girl of mass mg=47 kg is pulling a sled up a slippery slope. The coefficient of friction between the girl's boots and the slope is µs=0.160; the friction between the sled and the slope is negligible. The girl can pull the sled up the slope with a ≤ amax=0.015 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 = 6°.

If the sled is being pulled up the slope with a=amax, what is the mass of the sled?

mg = 47kg
µs=0.160
amax=0.015 m/s2
θ = 6°

Homework Equations

fnet=ma
Fs = Fnµs

The Attempt at a Solution

m(0.015m/s^2) = Fs - Mgsinθ
m(0.015m/s^2) = 73.3N + m(9.81m/s^2)sin6
m = 72.54kg

72.54kg-45kg = 27.5kg


i tried everything 27.5kg isn't the right answer what am i doing wrong someone please help i need to figure this out by tomorrow

 

[anathema]

Thank you for your post. After reviewing your attempted solution, I noticed that you have made a few errors in your calculations. Here is the correct solution to the problem:

First, let's draw a free body diagram of the girl and sled on the slope:

|Fn | |Fp|
| | | |
| Mg | |Mgsinθ|
| | | |
| Fg | |Fgsinθ|

Where Fn is the normal force, Fp is the pulling force exerted by the girl, Fg is the weight of the girl, and Fgsinθ is the component of the weight parallel to the slope. Since the sled has negligible friction, we only need to consider the forces acting on the girl.

Now, using Newton's second law, we can write the following equation:

Fnet = ma

Where Fnet is the net force acting on the girl, m is her mass, and a is her acceleration. We can break down the net force into its components:

Fnet = Fp - Fgsinθ - Fs

Where Fs is the force of friction acting on the girl. We can also express Fs in terms of the coefficient of friction and the normal force:

Fs = µsFn

Substituting this into our equation for Fnet, we get:

Fnet = Fp - Fgsinθ - µsFn

Now, using trigonometry, we can express Fgsinθ in terms of the girl's weight:

Fgsinθ = Mgsinθ

Substituting this into our equation for Fnet, we get:

Fnet = Fp - Mgsinθ - µsFn

Finally, we can substitute the values given in the problem into this equation:

Fnet = 0.015m/s^2 * 47kg - (47kg * 9.81m/s^2 * sin6°) - (0.160 * Fn)

We can also express Fn in terms of the girl's weight:

Fn = Mgsin6°

Substituting this into our equation, we get:

Fnet = 0.015m/s^2 * 47kg - (47kg * 9.81m/s^2 * sin6°) - (0.160 * 47kg * sin6°)

Simplifying this, we get: