A solid bar of length L = 1.32 m has a mass m1 = 0.751 kg. The bar is fastened by a pivot at one end to a wall which is at an angle θ = 35.0 ° with respect to the horizontal. (in other words, the angle between the wall and the beam is 35 degrees).The bar is held horizontal by a vertical cord that is fastened to the bar at a distance xcord = 0.725 m from the wall. A mass m2 = 0.722 kg is suspended from the free end of the bar.

Question

A solid bar of length L = 1.32 m has a mass m1 = 0.751 kg. The bar is fastened by a pivot at one end to a wall which is at an angle θ = 35.0 ° with respect to the horizontal. (in other words, the angle between the wall and the beam is 35 degrees).The bar is held horizontal by a vertical cord that is fastened to the bar at a distance xcord = 0.725 m from the wall. A mass m2 = 0.722 kg is suspended from the free end of the bar.
What is the vertical component of the force exerted by the wall onto the beam? Assume the positive direction is upwards. 
I know that the tension in the string is 20.2, and that the horizontal component is 0, but I'm having trouble creating the correction equation for the vertical component. My guess was Fy = m1g + mg2 - Tsin(55 degrees), but it turns out to be wrong. Help?

bobpursley · Answer

you "know" tension is 20.2 N? Sum moments about the hinge (m1*g*L/2+m2*g*L-T*L) sinTheta=0 L*g(m1/2+m2)=TL T= g(.751/2+.722) =9.81(1.09)=10.7 did I miss something?

Misaki · Answer

For tension in the cable, I did 7.51N (1.32/2) + 7.22 (1.32-0.775)= 9.25. Then I took that number and divided it by 1.32, which equals 7.01. I then divided 7.01 by sin(55 degrees), which gives me a Tension of 20.2.

Misaki · Answer

The force of gravity is also 10 N/kg for this problem.