Question

Tweedeldum and Tweedledee are carrying a uniform wooden board that is L = 3.00 m long and has a weight of w = 160 N . If Tweedledum applies an upward force of magnitude F1 = 60.0 N at the left end of the board, at what point and with what magnitude F2 of force does Tweedledee have to lift for the board to be carried?

Answer 1

100 N and 0.9 m from the center towards right.

Step-by-step explanation:

Given:

length of the wooden board, L = 3.00 m

weight of the wooden board, W = 160 N

`F_1 = 60.0 N` (at the left end)

`F_2 = ?`

Calculation:

In order to lift the board, the weight of the wooden board (acting downwards) must be equal to the applied upward force. So,

`W = F_1 +F_2\\160 N = 60.0 N + F_2\\ \Rightarrow F_2 = 100 N`

The net torque on the board should be zero. Consider center of mass at the center of the board i.e at = 1.5 m from the left end.

`(60.N) (-1.5m) + (100 N) (r) = 0\\ \Rightarrow r = 0.9 m` from the center towards right.

Answer 2

100 N force have to lift for the board to be carried

Step-by-step explanation:

Given data

L = 3.00 m

F1 = 60.0 N

w = 160 N

to find out

magnitude F2 of force

solution

we know that F1 and F2 sum is equal to weight of board

so that we can say

F1 + F2 = W

so

F2 = W - F1

put the value W and F1

F2 = 160 - 60

F2 = 100 N

so 100 N force have to lift for the board to be carried