In the figure below, a 36.0 kg uniform square sign, of edge L = 2.00 m, Is hung
ID: 1443809 • Letter: I
Question
In the figure below, a 36.0 kg uniform square sign, of edge L = 2.00 m, Is hung from a horizontal rod of length d_b = 3.00 m and negligible maw. A cable is attached to the end of the rod and to a point on the wall at distance d_v = 4.00 m above the point where the rod is hinged to the wall. What is the tension in the cable? What are the magnitude and direction of the horizontal component of the force on the rod from the wall? What are the magnitude and direction of the vertical component of the force on the rod from the wall? Did you a write A balanced-of-torques equation, using the hinge as the rotation axis for calculating torques? Did you apply the force due to the sign's weight midway between the sign's attachment points? Do you recall how to calculate a torque given a force's magnitude and angle? After you get the tension, did you apply A balance-of-forces equation horizontally and vertically?Explanation / Answer
First, 36 kg is about F = mg = 36 x 9.81 = 353.16 newtons.
Tension in the cable is very easy to find. Just balance moments on the rod about/around the hinge. Consider clockwise moment to be positive. Sum of the moments has to equal zero.
0 = 353.16/2 (1) + 353.16/2 (3) - Tv (3)
Tv = vertical force on the rod by the cable = vertical component of cable tension force = 235.44 newtons
Tension in cable = T = sqrt(Th^2 + Tv^2)
where Th and Tv are the horizontal and vertical components of T
Now we'll express Th and Tv in terms of T based on geometry. Think of it like this, based on just the angle of the cable, you can tell how much of T goes in the horizontal and how much goes in the vertical.
Th = 3/5 T
Tv = 4/5 T
235.44 newtons = 4/5 T
T = 294.3 N
Th = 3/5 (294.3 N) = 176.58 newtons
Thus, your answers are:
a) 235.44 newtons
b) 176.58 newtons
wall pushes the rod in compression to the right
c) vertical reaction at hinge = 353.16 newtons - 235.44 netwons = 117.72 newtons (upward)