Question

A square ABCD of 2m side is subjected to forces of 10N ,20N,30N,40N and 16√2N along AB ,AD,CB,CD andBD .find their resultant moment about the corner A.

Answer 1

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Answer 2

resultant moment about the corner A is 36 Nm

To Find:

A square ABCD of 2 m side is subjected to forces of 10 N, 20 N, 30 N, 40 N and 16√2 N along with AB, AD, CB, CD, and BD.

find their resultant moment about the corner A.

Explanation:

We know that,

moment = force * perpendicular distance

Therefore, force at AB, AD has perpendicular distance 0.

So moment due to force AB, AD is zero.

⇒Let the clockwise direction be positive.

Therefore, moment because of force CB = 30*2 = 60

Since it is anticlockwise, so it is negative.

Thus, Mcb = -60 Nm

moment due of force CD = 40*2 = 80

Since it is clockwise, so it is positive.

Thus, Mcd = 80 Nm

moment due of force BD = 16√2*(1/√2) = 16

Since it is clockwise, so it is positive.

Thus, Mbd = 16 Nm

Hence, total moment = Mcb+Mcd+Mbd = -60 + 80 + 16 = 36 Nm

Hence, resultant moment about the corner A is 36 Nm