The 60kg block is placed on the 30 degree incline against th

The 60-kg block is placed on the 30 degree incline against the spring and released from rest. The coefficient of static friction between the block and the incline is 0.39. (a) Determine the maximum and minimum values of the initial compression force in the spring (enter a positive number) for which the block will not slip upon release. (b) Calculate the magnitude and direction of the friction force acting on the block if the spring compression force C is 307 N. The friction force is positive if up the incline, negative if down the incline. Answers: (a) C_max = N C_min = N (b) F = N

Solution

Force normal to incline N = mg*cos 30 = 60*9.81*cos30 = 509.74 N

Static friction force = 0.39*N = 0.39*509.74 = 198.8 N

Force due to weight along the incline = mg*sin30 = 60*9.81*sin30 = 294.3 N

a)

To stop the downward movement, Cmin = 294.3 - 198.8 = 95.5 N

To stop the upward movement, Cmax = 294.3 + 198.8 = 493.1 N

b)

Net Upward force due to compression = 307 - 294.3 = 12.7 N

Since 12.7 < 198.8, the friciton force will be 12.7 N acting downward or -12.7 N.