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Question
Repeat part (a) of problem 6 if the push is applied horizontally and not parallel to the incline.
Solution
Using the free body diagram,
g = 10 m/s2, m = 2 kg, θ = 30 and μ = 0.2
R − mg cos θ − F sin θ = 0
⇒ R = mg cos θ + F sin θ (1)
and
mg sin θ + μR − F cos θ = 0
⇒ mg sin θ + μ(mg cos θ + F sin θ) − F cos θ = 0
⇒ mg sin θ + μmg cos θ + μF sin θ − F cos θ = 0
`=>F=(mg sintheta+mu mgcostheta)/(musintheta-costheta)(theta=30^circ)`
`=(2xx20xx(1/2)+0.2xx2xx10xxsqrt3/2)/(0.2xx(1/2)-(sqrt3/2)`
`13.464/0.76=17.7`N = 17.5 N
Therefore, while pushing the block to move up on the incline, the required force is 17.5 N.
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