A 2.00 kg block is held in equilibrium on an incline of angle θ = 45° - Don't Understand - HELP!?
A 2.00 kg block is held in equilibrium on an incline of angle θ = 45° by a horizontal force vector F applied in the direction shown in the figure below. If the coefficient of static friction between block and incline is μs = 0.300, determine the following.
Picture: http://www.webassign.net/sercp/p4-50.gif
(a) the minimum value of vector F
_____N
(b) the normal force exerted by the incline on the block
_____N
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Verified answer
Motive force Fm = mgsinθ = 2*10*0.707 = 14.14 N
Friction force Fr = mgcosθμs = 2*10*0.707*0.3 = Fm*0.3 = 4,242N
F =( Fm-Fr)/cosθ = 2*10*0.707*(1-0.3)/0.707 = 2*10*0.7 = 14.0 N
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