A 53.3 kg woman slides down a 35.0 degree hill with an acceleration of 4.10 m/s2. What is the friction force acting on the woman?

Answer 1

Let us try to find a solution to this problem without (unfortunately) using a body-diagram (which would make it a lot easier to visualize):

Given:

m = 53.3 kg.

g = 9.80 m./s²

θ = 35.0⁰

a = 4.10 m./s²

Find:

F_μ = ?? N.

Since the problem states the woman is on an incline, the vertical force is given by:

m · g

Therefore, the normal force, N, is given by:

N = m · g · cos (θ)

And the force along the ramp is given by:

F = m · g · sin (θ)

The sum of the forces along the incline is given by:

F - F_μ = ma

Solving for the force of friction:

1. F - F_μ = ma
2. F_μ = F - ma

Therefore, one finds that the force of fiction has a value of:

1. F_μ = F - ma
2. F_μ = m · g · sin θ - m · a
3. F_μ = ( 53.3 kg. ) [ ( 9.80 m./s² ) sin ( 35.0⁰ ) - ( 4.10 m./s² ) ]
4. F_μ ≈ 81.07191576 kg.·m·s^-2
5. F_μ ≈ 81.1 N.