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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 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.

Respuesta :

Answer:

T = 3600 lb

Explanation:

Given:

- coefficient of static friction @a u_a = 0.2

- coefficient of static friction @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 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 @ point A:

                      F_a = u_a*N_a = 4636.364*0.2

                      F_a = 927.2727 lb

- Take 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 sum of forces in 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 equation simultaneously:

                      T = 3600 lb , N_b = 2882 lb

                   

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