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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?

Respuesta :

Answer:

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

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

Answer:

100 N and 0.9 m from the center towards right.

Explanation:

Given:

length of the wooden board, L = 3.00 m

weight of the wooden board, W = 160 N

[tex]F_1 = 60.0 N[/tex] (at the left end)

[tex]F_2 = ?[/tex]

Calculation:

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

[tex]W = F_1 +F_2\\160 N = 60.0 N + F_2\\ \Rightarrow F_2 = 100 N[/tex]

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.  

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

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