Question

The 70 N force acts on the end of the pipe at B shown in Fig. determine a) the moment of this force about point A, and (b) the magnitude and direction of a horizontal force , applied at C, which produces the same moment. Take θ = 60° *

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Answer 1

Explanation:

Moment of force about point A will be given as

τ=Fcosθ×4+Fsinθ×2

For maximu τ,

dθ

dτ

=0

0=−4Fsinθ+2Fcosθ

tanθ=

2

1

Answer 2

The question is incomplete without a figure. So, the figure is given below.

The moment of this force at point A is 73.9 in a clockwise direction.

The magnitude and direction of a horizontal force applied at C is 82.2 N and the direction is from east to west.

Given:

Force, Fb = 70N acts on the end of the pipe at B.

From the figure given below a = 0.9 m, b = 0.3 m, c = 0.7 m.

To Find:

a) The moment of the force 70N about point A.

b) The magnitude and direction of a horizontal force applied at C, which produces the same moment as force Fb at A. Take θ = 60°

Solution:

a) We are required to find the moment at A

We know the moment is equal to Force×perpendiculardistance.

The force at B is resolved into two components

1) Horizontal component = Fb cos60°

2) Vertical component = Fb sin60°

The moment at A, Mₐ = F cos60°×a +  F sin60°×c

Mₐ = (70×1/2×0.9) + (70×√3/2×0.7)

     = (35×0.9) + (35×√3×0.7)

     = 31.5+42.4

     = 73.9 in clockwise direction (+)

Therefore, The moment of this force at point A is 73.9 in a clockwise direction.

b) We are required to find the direction and magnitude of the force applied at C which produces the same moment at A (Mₐ).

Mₐ = 73.9, a = 0.9 m

Then, Mₐ = Fc×a

Fc = Mₐ / a

    = 73.9 / 0.9

    = 82.2 N, the direction is from east to west.

Therefore, The magnitude and direction of a horizontal force applied at C is 82.2 N and the direction is from east to west.

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