So I have looked at several ladder example problems, (the most basic for statics), but was given a problem by my professor that has me a little stumped. This is day 2 of learning for me and I am caught up, heres the problem:
A 30.8 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 90o. If the beam is inclined at an angle of = 10.4o 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.)
I uploaded a pic so everyone can see the situation.
Well any basic trig knowledge, cos, sin, and tan.
The Attempt at a Solution
Nc= normal force by the cable
mg= mass times gravity
Fby-forcy by beam in y
Fbx= force by beam in x
I tried breaking it into letter variables first and solving that way instead of plugging in numbers the whole time.
I know I should set my Net force in all directions to zero.
I Set the force by the cable on the wall equal to the force of the beam in the x direction.
Then for the y direction the force of the beam (Fby)= mg
Then for torque I did T=-1/2Lcostheta(mg)+NcLsintheta
But I am really confused on how to set all of this up. Please help? I resulted with Nc=m/2*cottheta
But The number I entered gave me a wrong answer for the computer. What am I doing wrong?
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