Answer:
(M)_A = (M)_b = 76027.5 Nm
Explanation:
Step 1:
- We will first mark each weight from left most to right most.
Point: Weight:
G_3 W_g3 = 6 Mg*g
G_2 W_g2 = 0.5 Mg*g
G_1 W_g1 = 1.5 Mg*g
Load W_L = 2 Mg*g
Step 2:
- Set up a sum of moments about pivot point B, the expression would be as follows:
(M)_b = W_g3 * 7.5 + W_g2*4 - W_g1*9.5 - W_L*12.5
Step 3:
- Plug in the values and solve for (M)_b, as follows:
(M)_b = 9.81*10^3( 6* 7.5 + 0.5*4 - 1.5*9.5 - 2*12.5)
(M)_b = 9.81*10^3 * (7.75)
(M)_b = 76027.5 Nm
Step 4:
- Extend each line of action of force downwards, we can see that value of each force does not change also the moment arms of each individual weight also remains same. Hence. we can say that (M)_A = (M)_b = 76027.5 Nm