Magnitude of the horizontal component of the force of the wall acting on the bar?

A uniform horizontal bar of mass m=2.34kg and length L=2.02m is held by a frictionless pin at a wall. The opposite end of the strut is supported by a cord with tension T at an angle theta. A block of mass m2=4.68kg is hung from the bar at a distance of 3/4L from the pin. The acceleration of gravity is 9.8m/s2.


For the first part of the problem, I found the tension in the cord. If w=mg.. I found that T=2w/sintheta. I don't know if that helps for this problem but I thought I'd add it on here anyways.


Now, it wants to find the magnitude of the horizontal component of the force of the wall acting on the bar. (If theta=38.7 degrees). Answer in units of N.


I know the only horizontal force is T*costheta.. with T being the tension in the cord. I don't know where to go from there. Any suggestions would be appreciated!|||This is a problem in statics, that is the system is in equilibrium with no acceleration. The moment of all forces will add to 0. You first determine the forces acting on the bar at each distance from the pin. The must add up to zero.





The three forces are weight of the bar (W) which can be assumed to act at the midpoint L/2, the weight of the block (Wb) acting at 3/4L, and the vertical component of T (Tv) acting at L.





(1) Tv * L - W * L/2 - Wb * 3L/4 = 0





Since you know weight in newtons is mass * G, and you know the length of the bar L, the mass of the bar and the block, as well as G, the equation (1) has only one unknown, Tv, and is solvable.





(2) T = Tv / sin(theta)





Again, you know Tv from (1) and theta is given as 38.7 degrees, (2) can be solved.





The horizontal component Th of T is





(3) Th = T * cos(theta)





The horizontal force of the wall on the bar will be equal and opposite of Th.|||ok so . Equilibrium question the Forces all must cancel. So as you mentioned the horizontal components must all sum to = 0 . So then your Tcos(theta) + Fwall in x direction = 0. Therefore the Force of the horizontal component of the Wall is just the negative of the tension.





Fwallinx = -Tcos(theta)