A 20 kg wagon is released from rest from the top of a 15 m long plane, which is angled at 30° with the horizontal. Assuming there is friction between the ramp and the wagon, how is this frictional force affected if the angle of the incline is increased?
∑Fy = n - mgcos30° = 0
∑Fx = -ff (force of friction) + mg sin 30° = ma
up = +
down = -
→ = +
← = -
ff (force of friction) = μ (coefficient of friction) * n (normal force)
The Attempt at a Solution
I approached this problem by first finding the normal force when angle θ = 30°.
Therefore, ∑Fy = n-mgcos30° = 0
⇒n = mgcos30°
⇒n = 174 N
Then I solved ∑Fy = 0 when angle θ is increased, for example when θ = 60°
Therefore, ∑Fy = n-mgcos60° = 0
⇒n = mgcos60°
⇒n = 100 N
Then, I solved ∑Fx = mgsin30° - ff = ma when θ = 30°
⇒ff = mgsin30° - ma
⇒ff = 100 - 20a
Then, I solved ∑Fx = mgsin60° - ff = ma when θ = 60°
⇒ff = mgsin60° - ma
⇒ff = 174 - 20a
For example if I designate that "a" = 2 m/s^2, then ff = 100 - 20(2) ⇒60 N when θ = 30°
and ff = 174 - 20(2) ⇒ 134 N when θ = 60°. Therefore, when angle of incline increases, friction force increases.
However, my logic is wrong. The solution is that since ff = μ * n , and if angle of incline increases, friction force decreases, since ff = μ * 174 N when θ = 30° and ff = μ * 100 N when θ = 60°. I don't know why my approach is wrong...Any help would by greatly appreciated. Thanks.