| mg |
 |
| 2 sin(34°) |
An object at equilibrium has no net force acting on it, and has no net torque acting on it.
ΣF = 0
Στ = 0
A horizontal hinged rod is supported by a rope at one end. The rod has a mass of 1.4 kg, and there is an angle of 34° between the rope and the rod.
(a) What is the tension in the rope?
(b) What are the two components of the support force exerted on the rod by the hinge?
Start with a free-body diagram. If you aren't sure which way a force goes take a guess - if you guess wrong you'll just get a negative sign for that force.
Assume the rod is uniform so the force of gravity acts at the center of the rod. As usual, sum the forces in the x and y directions:
ΣFy = 0
Hy + T sin(34°) - mg = 0
ΣFx = 0
Hx + T cos(34°) = 0
There are too many unknowns - this is where torques come in. A sensible point to take torques about is one that one or more unknown forces go through, so they won't appear in the torque equation. Choosing the hinge as the point to take torques around eliminates both components of the hinge force. In this case, let's make clockwise positive.
Στ = 0
0.5 L mg - L T sin(34°) = 0
L, the length of the rod, cancels and we can solve for T.
| T | = |
|
= 12.3 N |
Substitute this back into the force equations to find the components of the hinge force:
Hx = T cos(34°) = 10.2 N
Hy = mg - T sin(34°) = 6.86 N