A 50kg sign for a local business is hung from a bar which meets a wall at right angles as shown in figure 1. The bar is 1.5 meters long, and it's mass may be neglected. The center of mass of th sign is 1.0 meters from the wall (but .50 meters from the tip of the cable) as indicated. A cable (we assume massless) extends from the end of the rod to the wall. Assume the rod is attached to the wall with a joint which permits the rod to rotate freely. The angle between the cable and bar is 30 degrees. What is the tension in the cable.
I know to get the mass you do MG which is 491N I am struggling on what to do since the mass is not hanging on the end but between two different lengths, how do I use these two lengths to solve my answer? I believe the answer is 650N
at one meter from the hinge at the wall you have 491 N down
at 1.5 meters from the wall you have T sin 30 up
moments about hinge at wall yield:
(T sin 30) * 1.5 = 491 * 1
T (1/2)(3/2) = 491
T = 491 * 4/3
To solve this problem, you can use the concept of torque equilibrium. The torque about the joint where the rod is attached to the wall should be balanced in order to keep the system in equilibrium.
First, calculate the torque exerted by the weight of the sign. The torque is the product of the weight (force due to gravity) and the perpendicular distance from the point of rotation (joint) to the line of action of the weight. In this case, the perpendicular distance is 1.0 meters since the center of mass of the sign is at a distance of 1.0 meters from the wall.
Torque due to the weight = (mass x gravitational acceleration) x perpendicular distance from the joint to the line of action of the weight
= (50 kg x 9.8 m/s^2) x 1.0 meters
= 490 N.m
Next, calculate the torque exerted by the tension in the cable. The magnitude of the torque due to tension can be found using the same formula, but this time the perpendicular distance is given by the entire length of the rod, which is 1.5 meters.
Torque due to tension = tension in the cable x perpendicular distance from the joint to the line of action of the tension
= T x 1.5 meters
Since the angle between the cable and the bar is 30 degrees, the perpendicular distance can be found using trigonometry. The perpendicular distance is given by the length of the rod multiplied by the sine of the angle.
Perpendicular distance = length of rod x sin(angle)
= 1.5 meters x sin(30 degrees)
= 0.75 meters
Now, we can write the equation for the torque equilibrium:
Torque due to weight = Torque due to tension
490 N.m = T x 0.75 meters
Solving for T:
T = 490 N.m / 0.75 meters
T ≈ 653.33 N
Therefore, the tension in the cable is approximately 653.33 N.