Beam hinged to wall with a cable (torque, etc)

[lizzyb]

Hi. I've tried to answer this questions four times and so far none of my answers have been correct. Any help is greatly appreciated!

A uniform 48 kg (m) beam at an angle of 24 degrees (theta) with respect to the horizontal has length of 38 m (L). It is supported by a pin and horizontal cable. The acceleration of gravity is 9.8 m/s^2 (g). What is the magnitude of the total force exerted by the pin on the beam?

So far:

$$\sum F_x = F_h - T = 0$$ horizontal component of pin - tension of cable
$$\sum F_y = F_v - mg = 0$$ vertical component of pin - mass of beam * gravity
$$\sum \tau = \sin \theta L T - \cos \theta L m g = 0$$

Does this look ok to you? For the torque, I'm trying to use the moment arm times the magnitude of the force.

Thank you.

 

[lizzyb]

The last equation should have been:
$$\sum \tau = \sin \theta L T - \cos \theta \frac{L}{2} m g = 0$$
and that took care of it.

 

[kyle8921]

Hi there,

Based on the information provided, your equations and approach seem correct. To find the magnitude of the total force exerted by the pin on the beam, you can use the Pythagorean theorem:

F_total = \sqrt{F_h^2 + F_v^2}

Where F_h is the horizontal component of the pin and F_v is the vertical component of the pin. This will give you the total force exerted by the pin on the beam, taking into account both the horizontal and vertical components.

As for the torque equation, it looks like you are using the right formula, which is the moment arm (in this case, the length of the beam) multiplied by the magnitude of the force (tension of the cable or weight of the beam). Just make sure to use the correct sign conventions for the angles and directions of the forces.

I hope this helps! Let me know if you have any further questions. Keep up the good work!