posted by Ana on .
A 600 N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long, uniform, 90-N rod. The left end the rod is supported by a hinge and the right end is supported by a thin cable making a 30.0° angle with the vertical.
(a) Find the tension, T, in the cable.
(b) Find the horizontal and vertical components of force exerted on the left end of the rod by the hinge. (Take up and to the right to be the positive directions.)
horizontal component_____ N
We need to know how far from the wall the center of mass of the sign is.
It wouldn't allows me to submit picture
OK, but tell me from the picture how far the middle of the sign is from the wall. Is the middle of the sign in the middle of the rod (3 meters from wall) or is it all the way out (4 meters out)
sign is away from wall
Ok, will assume it goes out to the end
take moments about the hinge.
T * 6 * cos 30 = 5.2 T counterclockwise
600*4 + 90*3 = 2670 clockwise
so T = 2670/5.2 = 513 Newtons tension
forces right on rod = 0 = Fwallhorizontal - Tsin 30
So Fwall horizontal = 513/2 = 257 N
forces up on rod = Fupwall +513 cos 30
forces down on on rod = 690
Fupwall = 690 - 444 = 246 N
It did work. thank you so much
You are welcome.