1. Consider 7.0 kg mass pulled with a force of 60 N up an incline with an angle of 15º. The coefficient of friction between the surfaces is 0.18.
Given: m =7.0 kg; θ = 15°; Fpuill = 60 N; μ = …… ..Fg = 68.6 N
For ANY object on an incline, Fg can be broken into two components:
F|| (Fg sinӨ) ▬ always parallel to and down the incline
FN (Fg cosӨ) ▬ always perpendicular to the incline a
Draw the free body diagram: FN
F = ma
F = Fpull – f - F||
ma = Fpull – f - F||
Fpull
F|| f Fg

2. Determine F|| and FN from Fg:
FN = Fg cosӨ
F|| = Fg sinӨ
F|| = (68.6) sin(15º)
= 17.8 N
FN = (68.6) cos(15º)
= 66.3 N
Determine friction from FN and µ:
ma = Fpull – f - F||
7.0 (a) = 60 – 11.9 – 17.8
Substitute in knowns and solve for a
f =μFN
f = (.18)(66.3)
f = 11.9 N
a = 4.33 m/s2

3. FTx= FT cosӨ = 10.2 N FTy = FT sinӨ = 6.4 N
Consider a 3 kg block pulled along a horizontal
surface at constant velocity by a force of 12N at an angle of 32º. What is the coefficient of
friction between the block and the surface?
Draw the free body diagram: FN
Break FT into its components FT
32º FT f
FTy
32º Fg
FTx
a = 0
FTx= FT cosӨ = 10.2 N FTy = FT sinӨ = 6.4 N

4. Recall: The y-component of the pull reduces the normal force (the y-component of a push would increase the normal force)
F = FTx – f
F = ma
ma = FTx – f
FN = Fg - FTy
FN =
FN = 23.0 N
(3)(0) = 10.2 – f
f = 10.2 N
f =μFN
10.2 = μ(23.0)
µ = 0.44