The coefficients of static friction at a and b are T = 3600 lb and N_b = 2882 lb.
Given:
Coefficient of static friction at a u_a = 0.2
Coefficient of static friction at b u_b = 0.3
Weight of the loaded bin W = 8500 lb
Find:
Find the force in the cable needed to begin the lift.
Solution:
Draw the forces on the diagram. see attachment.
Take the sum of moments about point B as zero:
(M)_b = W*12 - N_a * 22 = 0
N_a = W*12 / 22 = 8500*12 / 22
N_a = 4636.364 lb
Compute friction force F_a at point A:
F_a = u_a*N_a = 4636.364*0.2
F_a = 927.2727 lb
Take the sum of moments about point A as zero:
-W*10 - F_b*sin(30)*22+ 22*N_b*cos(30) + 22*T*sin(30) = 0
Where, F_b = u_b*N_b = N_b*0.3
Hence, -85000 - 3.3*N_b + 11sqrt(3)*N_b + 11 T = 0
15.753*N_b + 11*T = 85000 ...... (1)
Take the sum of forces in the x-direction equal to zero:
T*cos(30) - N_b*sin(30) - u_b*N_b*cos(30) - F_a = 0
T*cos(30) - 0.75981*N_b = 927.2727 ..... (2)
Solve two equations simultaneously:
T = 3600 lb , N_b = 2882 lb
Refer to this complete question:
The winch on the truck is used to hoist the garbage bin onto the bed of the truck. If the loaded bin has a weight of 8500 lb and a center of gravity at G, determine the force in the cable needed to begin the lift. The coefficients of static friction at A and B are MuA = 0.2 And MuB = 0.3 respectively. Neglect the height of the Support at A.
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