Asked by anonymous
A horizontal force F~ is applied to a block of mass m = 1 kg placed on an inclined at θ = 30◦ plane. The coefficients of static and dynamic friction are µs = 0.3 and µ = 0.2, respectively.Find F such that the block is moving up the slope with a constant speed.
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Answered by
Anonymous
You really mean F is horizontal?
If so F up slope = F cos 30 = 0.866 F
weight component down slope = m g sin 30
normal force = F sin 30 + m g cos 30 = F/2 + 0.866 m g
so friction force down slope= 0.2 ( F/2 + 0.866 m g)
so force up = force down if not accelerating
0.866 F = mg /2 + 0.2 ( F/2 + 0.866 m g)
If so F up slope = F cos 30 = 0.866 F
weight component down slope = m g sin 30
normal force = F sin 30 + m g cos 30 = F/2 + 0.866 m g
so friction force down slope= 0.2 ( F/2 + 0.866 m g)
so force up = force down if not accelerating
0.866 F = mg /2 + 0.2 ( F/2 + 0.866 m g)
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