Force of a hinge on a hinged beam

[Becca93]

Homework Statement

A 26.6 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the cable is 90°. If the beam is inclined at an angle of theta=13.3° with respect to horizontal, what is the horizontal component of the force exerted by the hinge on the beam? (Use the 'to the right' as + for the horizontal direction.)

Hint: The Net torque and the Net Force on the hinge must be zero since it is in equilibrium.

What is the magnitude of the force that the beam exerts on the hinge?

(Image attached)

The Attempt at a Solution

I already knew that the net force and net torque would be zero, so I set clockwise and counter clockwise torque equal to each other

τcw = τccw
Fdsinθ = Fdsinθ

And this is where I ran into trouble. Length of the beam is never given. I'm not really sure what angles to use where, and while I know that the force the beam exerts on the wall/hinge is equal and opposite to what the wall/hinge exerts on the beam, I'm not sure how to find it.

To find the magnitude of force in the second half of the problem, you would just take magnitude = √(Fx^2 + Fy^2), correct?

Any advice?

 

[PhanthomJay]

It is best to first find the tension force in the cable by summing moments about the hinge.. the beams weight acts at midpoint and moment is force times perpendicular distance.. Your sohcahtoa is off. Try again.

 

[Becca93]

It is best to first find the tension force in the cable by summing moments about the hinge.. the beams weight acts at midpoint and moment is force times perpendicular distance.. Your sohcahtoa is off. Try again.

Okay, I understand that I need to use cos instead of sin, but I still don't understand what to do without the length of the beam.

 

[PhanthomJay]

Sum moments of the forces about the hinge. aybe you don't need to know thw beam length.

 

[asteriatic]

I would approach this problem by first defining the variables and forces involved. In this case, we have a 26.6 kg beam attached to a wall with a hinge, supported by a cable, and inclined at an angle of 13.3° with respect to horizontal. The forces acting on the beam are the weight of the beam (mg), the tension in the cable (T), and the force exerted by the hinge on the beam (Fh).

To solve for the horizontal component of the force exerted by the hinge (Fhx), we can use the fact that the net torque and net force on the hinge must be zero in equilibrium. This means that the sum of the clockwise and counterclockwise torques must be equal, and the sum of the horizontal and vertical forces must be equal to zero.

Using the given angle of 90° between the beam and the cable, we can see that the horizontal component of the tension in the cable (Tx) is equal to the weight of the beam (mg). Therefore, we can write the following equations:

τcw = τccw
Fhdsin13.3° = mgdsin90°
Fh = mgcos13.3°

Setting this equal to the horizontal component of the tension, we get:
Fhx = Tx = mgcos13.3°

To find the magnitude of the force that the beam exerts on the hinge, we can use the fact that the force exerted by the hinge on the beam is equal and opposite to the force exerted by the beam on the hinge. Therefore, the magnitude of the force exerted by the beam on the hinge is also mgcos13.3°.

To summarize, in order to find the horizontal component of the force exerted by the hinge on the beam, we used the fact that the net torque and net force on the hinge must be zero in equilibrium. And to find the magnitude of the force exerted by the beam on the hinge, we used the fact that these two forces are equal and opposite.