Question

The diagram in figure 1.29 shows two forces F1= 5N and F2=3N acting at points A and B of a rod pivoted at point O such that OA=2m and OB=4 m.

Calculate:
(1)the moment of force F1 about O
(2)the moment of force F2 about O
(3) total moment of the two forces about O​

Answer 1

(1) The moment of force F1 about O is 10 N-m

(2) The moment of force F2 about O is 30 N-m

(3) Total moment of the two forces about O is 20 N-m

Explanation:

Given

$F_1=5$ N

$F_2=3$ N

The moment of force $F_1$ will be anticlockwise and that of $F_2$ will be clockwise

Taking anticlockwise moment to be positive and clockwise moment to be negative

We know that

Moment = Force × Perpendicular Distance

The moment of force $F_1$

$\tau_1=F_1\times OA$

$\implies \tau_1=5\times 2$ N-m

$\implies \tau_1=10$ N-m

Moment of force $F_2$

$\tau_2=F_2\times OB$

$\implies \tau_2=3\times 4$ N-m

$\implies \tau_2=30$ N-m

Net moment

$\tau=\tau_1-\tau_2$

$\implies \tau=10-30$

$\implies \tau=-20$ N-m

Negative sign indicates clockwise moment.

Hope this helps.

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Answer 2

Answer:

Explanation:F1 = 5N , 2m

F2 = 3N , 4m

1. Moment of f1 = force ×oa

= 5×2

=10Nm(anticlockwise)

2. Moment of f2 = force × on

=3×4

= 12Nm( clockwise)

3. Total moment of the two force about O =

Moment of f2 - moment of f1

=12-10

=2Nm (clockwise)