Answered

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A 2 kg block is in equilibrium on a 36 degree incline. What is the normal force acting on the block?

Sagot :

Trancescient

15.9 N

Explanation:

Let's assume that the downward direction on the inclined is the (+)-direction. Since the block is in equilibrium, the x-component of its weight is pointing in the +x-direction and the frictional force [tex]f_s[/tex] is pointing up the incline. So the net force acting parallel to the incline can be written as

[tex]mg\sin36 - f_s = 0 \Rightarrow \mu N = mg\sin36[/tex]

where N is the normal force.

The net force perpendicular to the incline can be written as

[tex]N - mg\cos36 = 0 \Rightarrow N = mg\cos36[/tex]

or

[tex]N = (2\:\text{kg})(9.8\:\text{m/s}^2)\cos36 = 15.9\:\text{Newtons}[/tex]

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