The block A and attached rod have a combined mass of 60 kg and are confined to move along the 60° guide under the action of the 800-N applied force. The uniform horizontal rod has a mass of 20 kg and is welded to the block at B. Friction in the guide is negligible. Compute the bending moment M exerted by the weld on the rod at B

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Step 1 of 4Draw the following free body diagram:Step 2 of 4Consider the equilibrium of forces along X-direction, parallel to the applied force 800 \mathrm{~N}.80(1) m g \sin \theta-\maxHere, \mathrm{m} is the combined mass of the block and rod, and a is the acceleration of block along the \mathrm{X} direction.Substitute 60 \mathrm{~kg} for \mathrm{m}, and 9.8 \mathrm{~min} for \mathrm{g}.800-(60 \times 9.81) \sin 60^{\circ}=60 aa=4.84 \mathrm{~m} / \mathrm{s}^{2}Step 3 of 4To calculate the bending moment, consider only the rod and draw the free body diagram.Step 4 of 4Consider the equilibrium of moment at \mathbf{B}.Here, \mathrm{d} is the horizontal distance between the center of mass of the rod and end \mathrm{B}, \mathrm{M} is the bending moment exerted by the weld on \operatorname{rod} \mathrm{B}, and m is the mass of the rod.Substitute 20 \mathrm{~kg} for, and 0.7 \mathrm{~m} for \mathrm{d}.1.20 \log \times 0.720 \times 4.84 \times 0.7 \sin 60^{\circ}M-137.3=58.681 /=195.97 \mathrm{~N} \cdot \mathrm{min}Hence, the bending moment exerted by the weld on the rod is ...