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:
ΣF_{x} = F cos θ − Nμ = ma 
(1) 
ΣF_{y} = 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: 

(3) 
For the maximum force F that can be applied so that the mass m will still be at rest we have:
f_{max} = N μ_{s} = (m g − F sinθ) μ_{s}  (4) 
 (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: f_{max} < F cosθ.
Because 7.2 < 25.98 the mass will move and the acceleration is according to equation (3).
v_{t} = v_{0} + 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 μ_{k} / m = −g μ_{k} = − 9.8 * 0.2 = − 1.96 m/s^{2}
And the time until mass m will stop is according to the equation: v_{t} = v_{0} + a t where v_{t} = 0 and v_{0} = 38.5 m/s
t = − v_{0} / a = − 38.5 / − 1.96 = 19.64 sec
