Question

The block A and attached rod have a combined mass of 50 kg and are confined to move along the guide under the action of the 796-N applied force. The uniform horizontal rod has a mass of 15 kg and is welded to the block at B. Friction in the guide is negligible.
Required:
Compute the bending moment M exerted by the weld on the rod at B. The bending moment is positive if counterclockwise, negative if clockwise.

Answer 1

Answer:

The bending moment is 459.16 N.m

Explanation:

From the given information;

Let's assume that the angle is 66°

Then, the free body diagram is draw and attached in the file below.

Now, the calculation of the acceleration from the first part of the free body diagram is:

$\sum F_x = ma_x \\ \\ 796 - 50(9.81) sin 66=50a \\ \\ 796 - 448.094 = 50 a \\ \\ a = \dfrac{347.906}{50} \\ \\ a = 6.96 \ m/s^2$

Bending moment M:

From the second part of the diagram:

$\sum M_B = mad \\ \\ M - (15 \times 9.81) (1.5) = (25 \times 6.96)(1.5 sin 66) \\ \\ M - 220.725 = 238.435 \\ \\ M = 238.435 + 220.725 \\ \\ \mathbf{M = 459.16 \ N.m}$