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Explanation:
a) Here is the free-body diagram. Note that I included the components of the weight mg (shown in dotted arrows) for use in the other parts of the problem.
b) The component of the weight parallel to the plane (shown in the diagram as a dotted arrow along the x-axis) is [tex]mg\sin15[/tex] and it is equal to
[tex]mg\sin15 = (25\:\text{kg})(9.8\:\text{m/s}^2)\sin15 = 63.4\:\text{N}[/tex]
c) Applying Newton's 2nd law to the y-axis, we can write
[tex]y:\;\;\;N - mg\cos15 = 0 \Rightarrow N = mg\cos15[/tex]
[tex]N = (25\:\text{kg})(9.8\:\text{m/s}^2)\cos15 = 236.7\:\text{N}[/tex]
d) The component of the weight mg into the plane is the same as the normal force, hence it's also 236.7 N.
e) To solve for the coefficient of friction, we apply Newton's 2nd law to the x-axis:
[tex]x:\;\;\;mg\sin15 - F_f = 0[/tex]
[tex]\Rightarrow F_f = mg\sin15\;\;(2)[/tex]
where [tex]F_f[/tex] is the frictional force defined as [tex]F_f = \mu N[/tex] so we can use Eqn(1) on Eqn (2) to write
[tex]\mu (mg\cos15) = mg\sin15[/tex]
Solving for [tex]\mu,[/tex] we get
[tex]\mu = \dfrac{\sin15}{\cos15} = \tan15 = 0.27[/tex]
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Rewritten by : Barada
