Question

 A 100-lb load is suspended by two chains in a room. The angle between each and the horizontal ceiling is 45°. What is the magnitude of the force each chain must be support?

Answer 1

Answer:

Magnitude of force on each chain = 314.58 N

Explanation:

Refer the given figure to see the arrangement.

We have Vertical force of 100 lb mass downward.

That is mass = 100 lb = 45.35 kg

Weight = 45.35 x 9.81 = 444.88 N

On resolving F , we will have 2Fcos45 in upward direction.

Equating

         2Fcos45 = 444.88 N

          F = 314.58 N

Magnitude of force on each chain = 314.58 N

       

Answer 2

Magnitude of force on each chain suspended with 100 ib load in a room, must be support is 314.54 N.

What is static equilibrium state of hanging load?

If the load, is hanging on chain or rope, and is in the equilibrium state, then the sum of all the horizontal component of it must is equal to zero.

Given information-

The load suspended by two chains in a room is 100-ib.

The angle between each chain and the horizontal ceiling is 45°.

The image attached below shows the given situation.

First convert the 100-ib into unit of N as,

$F=100\rm ib=100\times4.4482\rm N\\F=444.82N$

As the load suspended by two chains in a room is 100-ib and the angle between each chain and the horizontal ceiling is 45°. Thus, the vertical component of the,

$\sin (45)=\dfrac{F_y}{444.82}\\F_y=314.54\rm N$

Hence, the magnitude of the force each chain must be support is 314.54 N.

Learn more about the equilibrium state here;

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