Question #278875

A 12N block rests on a plane inclined 15 degrees. If the block is not moving what is the force of friction acting on that block?

Expert's answer


Second Newton's law in projection on x-axis:


mgsin⁡α=Ffrmg\sin\alpha = F_{fr}

where mg=15Nmg = 15N is the weight of the block, α=15°\alpha=15\degree is the angle and FfrF_{fr} is the frictional force. Thus, obtain:


Ffr=15⋅sin⁡15°≈3.9NF_{fr}=15\cdot \sin15\degree \approx 3.9N

Answer. 3.9 N.


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