A 2.00 kg block is held in equilibrium on an incline of angle ? = 75° by a horiz

Question

A 2.00 kg block is held in equilibrium on an incline of angle ? = 75° 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.
(a) the minimum value of vector F
N

(b) the normal force exerted by the incline on the block
N

Explanation / Answer

Let: F be the horizontal force applied, R be the normal reaction of the plane on the block, S be the force of friction acting up the plane, m be the mass of the block, t be the angle of inclination of the plane to the horizontal, g be the acceleration due to gravity. Resolving parallel and perpendicular to the plane: F cos(t) + S = mg sin(t) ...(1) F sin(t) + mg cos(t) = R ...(2) The relationship between S and R is: S = mg[ (sin(t) - u cos(t)) / (cos(t) + u sin(t)) ] F >= 2.00 * 9.81[ (sin(75) - 0.300 cos(75)) / (cos(75) + 0.300 sin(75)) ] F >= 31.8 N. (b) From (2): R = F sin(75) + 2.00 * 9.81cos(75) For the minimum value of F: R = 35.8 N.

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