Math, asked by sweet33mona, 11 months ago

Fig. 1.31 shows two forces each of magnitude 10 N
acting at points A and B at a separation of 50 cm, in
opposite directions. Calculate the resultant
moment of the two forces about the point (i) A, (ii) B
and (iii) O situated exactly at the middle of the two
forces.
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Answers

Answered by AditiHegde
204

Fig. 1.31 shows two forces each of magnitude 10 N  acting at points A and B at a separation of 50 cm, in  opposite directions.

Given,

Force at A in upward direction = F = 10 N

Force at B in downward direction = F = 10 N

Distance between the points A and B = 50 cm = 0.5 m

(i) The resultant  moment of the two forces about the point A

= 10 N × 0.5 m

= 5 Nm in clockwise direction.

(ii) The resultant  moment of the two forces about the point B

= 10 N × 0.5 m

= 5 Nm in clockwise direction.

(iii) O situated exactly at the middle of the two  forces.

At point O, the perpendicular distance from either of the forces F = 10 N is 0.5/2 = 0.25 m.

The moment of the two forces at the point A about O

= 10 N × 0.25 m

= 2.5 Nm in clockwise direction.

The moment of the two forces at the point B about O

= 10 N × 0.25 m

= 2.5 Nm in clockwise direction.

Therefore, the moment of the two forces about O

= 2.5 Nm + 2.5 Nm

= 5 Nm in clockwise direction.

Answered by akankshasarki
62

Answer:

Step-by-step explanation:

As you can see that the movement is clockwise.

i) Moment of force A:  10N×0.5m=5Nm clockwise

ii) Moment of force B:

10N× 0.5m=5Nm clockwise

iii) In this part of the question we have to be clear that the answer is asked about the perpendicular distance between A and B.

The perpendicular distance between = 0.5/2=0.25

Now,

Moment of force A:

10N×0.5m=2.5Nm clockwise

Moment of force B:

10N× 0.5m=2.5Nm clockwise

The  moment of force exact at the middle of the two forces:

    2.5Nm+2.5Nm= 5Nm clockwise

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