Please I need help with this problem, the correct answer is 28.5 lb of tension in the cord AB. I tried many ways to solve it with no success. I need the correct free body diagram, equations, and procedure.
The picture of the problem is in the l;ink below. Thanks in advance for your efforts.
http://i1295.photobucket.com/albums/b640/Mdiazelecthrolite/2013-03-31143402_zps5e469b99.jpg
Mark P
Favorite Answer
There are 4 relevant forces. Draw all on your diagram
There are two forces at the lower bottom of the bar:
1. Vertical force of support (F_a)
2. Horizontal force to the right - tension in the rope ( F_b)
3. The third force is the weight (W = 86 lb) at the center of the bar, pointing straight down)
4. There is a reaction force (R) at the point of support (10 ft above the floor)
Once you have the forces on your diagram, we can solve the problem.
Solution:
The object is in equilibrium, therefore:
A) the sum of vertical forces is zero: F_a + R_y - W =0
B) the sum of horizontal forces is zero: F_b -R_x = 0
C) clockwise torque = counterclockwise torque: W*(arm_w) = R*(arm_R) ==>
86*8*cos60 = R* (10/cos30)
As you see the third equation (step C) can be solved for the reaction force R.
R = 29.8 lb
From step B, we conclude: F_b = R_x = (29.8)(cos30) = 25.8 lb