Physics questions regarding table and force?

The sum of the forces which the legs support is equal to the sum of the weights. Let’s determine the weight of the rod and block.

Rod = 10 * 9.8 = 98 N, Block = 2 * 9.8 = 19.6 N

Total weight = 117.6 N

Fa + Fb = 117.6

The clockwise torque is equal to the counter clockwise torque. Let the pivot point be at A. The weight of the rod and the weight of the block produce clockwise torque. The upward force at B produces counter clockwise torque.

Torque = Force * distance from the pivot point

The weight of the rod is 2.5 meters from A.

Clockwise torque = 98 * 2.5 = 245

The weight of the block is 4.25 meters from A.

Clockwise torque = 19.6 * 4.25 = 83.3

Total clockwise torque = 328.3

The upward force at B is 5 meters from A.

Clockwise torque = Fb * 5

Fb * 5 = 328.3

Fb = 65.66 N

Fa = 117.6 – 65.66 = 51.94 N

FBD of the problem

Na ............. ..... Wr ........ Wb ........... Nb

↑ ....... ........ ......↓ ......... ...↓ ........ ......↑

A________ cg-rod ____ block _______ B

|----- ----------- 5m ------------- ------------|

|------- 2.5 m ---- |

| ------------ 4.25 m --------- | - 0.75 m --|

Note: The CG of the rod is 1/2 the length.

Sum of the moments about A = 0 <---- No rotation and CCW = +

0 = Nb*5m - Wr*2.5m - Wb*4.25 m

Nb = (Wr*2.5 m + Wb*4.25m) / 5m

Nb = (10 kg * 9.81 m/s^2 * 2.5 m + 2 kg * 9.81 m/s^2 * 4.25 m) / 5 m

Nb = 65.7 N

Sum of the forces in the vertical direction = 0 <----- No acceleration and up is positive

0 = Na + Nb - Wr - Wb

Na = Wr + Wb - Nb

Na = 10kg * 9.81 m/s^2 + 2kg * 9.81 m/s^2 - 65.7 N

Na = 52.0 N