Question

The uniform boom shown below weighs 700 N, and the object hanging from its right end weighs 400 N. The boom is supported by a light cable and by a hinge at the wall. Calculate the tension in the cable and the force on the hinge on the boom. Does the force on the hinge act along the boom?

Answer 1

Answer:

Explanation:

I THINK THIS IS THE ANSWER  AS I HAVE NOT STUDIED THIS BEFORE

Answer 2

Answer:

Tension in the cable = 434N

Explanation:

From the diagram, we have created components of tension.

As system is in equilibrium:

Equating vertical components of force:

$Tsin(45^o)+F_2=400$

Equating horizontal components of force:

$Tcos(45^o)=F_1$

Equating torque on boom:

Say, length of boom in unity.

Torque = $r.F.sin(\theta)$

where, $r$ is the distance between hinge and force.

$F$ is the magnitude of force.

$\theta$ is the angle between $r$ and $F$.

$1.T.sin(65^o)=1.400.sin(80^o)\\\implies T = \frac{400 sin(80^o)}{sin(65^o)}=434N$

Yes, the force on the hinge act along the boom. This force is equal to the vector product of $F_1$ and $F_2$.

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