A body whose mass is m is resting on a horizontal surface whose friction coefficient is μ, a force F is implied on the mass at an angle of θ degree. Find the acceleration of mass m and the maximum force that the mass m will still be at rest.
From the forces diagram we can derive the equilibrium equations:
ΣFx = F cos θ − Nμ = ma |
(1) |
ΣFy = N + F sin θ − mg = 0 |
(2) |
From eq. (2) |
N = mg − F sin θ |
Substitute N into eq. (1) |
F cos θ − mgμ + Fμ sin θ = ma |
Solving for a we get: |
![Acceleration](/CalcPhysics/Friction/EQ/eq2.png) |
(3) |
For the maximum force F that can be applied so that the mass m will still be at rest we have:
fmax = N μs = (m g − F sinθ) μs | (4) |
![Maximum F for rest condition](/CalcPhysics/Friction/EQ/eq2a.png) | (5) |
A force of 30 N is applied to a mass of 3 kg for 5 sec at an angle of 30 degree, if the static and kinetic frictions are 0.5 and 0.2 respectively. Find the acceleration and the time needed for mass m to stop after force F is eliminated.
First check if the mass will move by the condition: fmax < F cosθ.
Because 7.2 < 25.98 the mass will move and the acceleration is according to equation (3).
vt = v0 + a t = 0 + 7.7 * 5 = 38.5 m/s
The new free body diagram after eliminating force F is at left.
According to the second law of motion F = m a (F now is only the friction force) and f = − m a (the minus sign is because of the opposite directions of a and f).
a = − f / m = − m g μk / m = −g μk = − 9.8 * 0.2 = − 1.96 m/s2
And the time until mass m will stop is according to the equation: vt = v0 + a t where vt = 0 and v0 = 38.5 m/s
t = − v0 / a = − 38.5 / − 1.96 = 19.64 sec
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