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GenaCL600 [577]
3 years ago
10

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.
Physics
1 answer:
Luba_88 [7]3 years ago
6 0

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}

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