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Need help with part b); please show step by step A symmetrical table of height 0.790...

Need help with part b); please show step by step

A symmetrical table of height 0.790 m, length 1.45 m, and weight 429 N is dragged across the floor by a force applied to its front edge. The force is directed to the right and upward and makes an angle of 30.0° with the horizontal.

(a) Find the minimum force necessary to drag the table across the floor. The coefficient of sliding friction between table and floor is 0.380.

154.37 N

(b) Calculate the normal force and the frictional force on each leg.

front legs: normal force

frictional force

back legs: normal force

frictional force

0 0
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Answer #1

The free body diagram is shown belo ext For the right leg, the torque due to weight of the table is, 1.451 +0.792 -(429)(1.07

Torque due to the frictional force, , -F (0.79) For the left leg, this is same as rights. Therefore, r, - F, (0.79) Now for

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