A uniform horizontal 200 N beam AB of length L has two weights hanging from it, 300 N at L3 from end I and 400 N at 3L4 from the same end. What single additional force acting on the beam will produced equilibrium?

The magnitude of single force should be200 + 300 + 400 = 900 N∑ τA = 0⇒300×L3+200×L2=400×3L4 = 900x⇒x = 59L = 0.555L = 0.56L

A uniform horizontal 200 N beam AB of length L has two weights hanging from it, 300 N at L3 from end I and 400 N at 3L4 from the same end. What single additional force acting on the beam will produced equilibrium?

The magnitude of single force should be200 + 300 + 400 = 900 N∑ τA = 0⇒300×L3+200×L2=400×3L4 = 900x⇒x = 59L = 0.555L = 0.56L

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A uniform horizontal 200 N beam AB of length L has two weights hanging from it, 300 N at $\frac{L}{3}$ from end I and 400 N at $\frac{3 L}{4}$ from the same end. What single additional force acting on the beam will produced equilibrium?

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900 N, in the middle

900 N, at 0.4L from A

900 N, at 0.46L from A

900 N, at 0.56L from A

answer is D.

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Detailed Solution

The magnitude of single force should be

$200 + 300 + 400 = 900 N$

$\sum_{} τ_{A} = 0$

$\Rightarrow 300 \times \frac{L}{3} + 200 \times \frac{L}{2} = 400 \times \frac{3 L}{4} = 900 x$

$\Rightarrow x = \frac{5}{9} L = 0.555 L = 0.56 L$

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A uniform horizontal 200 N beam AB of length L has two weights hanging from it, 300 N at L3 from end I and 400 N at 3L4 from the same end. What single additional force acting on the beam will produced equilibrium?

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A uniform horizontal 200 N beam AB of length L has two weights hanging from it, 300 N at $\frac{L}{3}$ from end I and 400 N at $\frac{3 L}{4}$ from the same end. What single additional force acting on the beam will produced equilibrium?