A solid bar of length L = 1.30 m has a mass m1 = 0.893 kg. The bar is fastened by a pivot at one end to a wall which is at an angle è = 33.0 ° with respect to the horizontal. The bar is held horizontal by a vertical cord that is fastened to the bar at a distance a distance xcord = 0.635 m from the wall. A mass m2 = 0.681 kg is suspended from the free end of the bar. Find the the vertical component of the force exerted on the bar by the wall. (Take the upward direction to be positive.)
A solid bar of length L = 1.30 m has a mass m1 = 0.893 kg. The bar is fastened by a pivot at one end to a wall which is at an angle è = 33.0 ° with respect to the horizontal. The bar is held horizontal by a vertical cord that is fastened to the bar at a distance a distance xcord = 0.635 m from the wall. A mass m2 = 0.681 kg is suspended from the free end of the bar. Find the the vertical component of the force exerted on the bar by the wall. (Take the upward direction to be positive.)
I can't quite get the picture of a wall being 33 degrees from the horizontal. Perhaps you omitted some important words.
To find the vertical component of the force exerted on the bar by the wall, we need to consider the torques acting on the bar.
First, let's determine the torque due to the weight of the bar itself. The weight of the bar can be calculated using the formula:
Weight = mass * gravity
Given that the mass of the bar is m1 = 0.893 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the weight of the bar.
Weight of the bar = 0.893 kg * 9.8 m/s^2 = 8.7564 N
The weight of the bar can be considered as acting at its center of mass, which is located at the midpoint of the bar.
Next, let's determine the torque due to the weight of the suspended mass (m2 = 0.681 kg). The torque due to the weight can be calculated using the formula:
Torque due to weight = weight * distance from the pivot
Since the suspended mass is located at the free end of the bar, the distance from the pivot to the suspended mass is equal to the length of the bar (L = 1.30 m). Therefore, the torque due to the weight of the suspended mass is:
Torque due to weight = 0.681 kg * 9.8 m/s^2 * 1.30 m = 8.3754 N·m
Now, let's determine the torque due to the tension in the vertical cord that holds the bar horizontal. The torque due to the tension can be calculated using the formula:
Torque due to tension = tension * distance from the pivot
The tension in the cord can be found by considering the equilibrium of forces in the vertical direction. The sum of the vertical forces must add up to zero, since the bar is held horizontal. Therefore, the vertical component of the force exerted on the bar by the wall must be equal to the sum of the weights of the bar and the suspended mass. Let's call the tension in the cord T.
Vertical component of force exerted on the bar by the wall = weight of the bar + weight of the suspended mass
Vertical component of force exerted on the bar by the wall = (0.893 kg * 9.8 m/s^2) + (0.681 kg * 9.8 m/s^2)
The distance from the pivot to the cord attachment point is given as xcord = 0.635 m. Therefore, the torque due to the tension in the cord is:
Torque due to tension = T * 0.635 m
Now, considering that the bar is in rotational equilibrium (i.e., the net torque is zero), we can equate the torques due to the weight and the tension. Therefore, we have:
Torque due to weight of the bar + Torque due to weight of the suspended mass = Torque due to tension in the cord
In other words:
8.7564 N * (1.30 m/2) + 8.3754 N·m = T * 0.635 m
Simplifying:
4.3782 N·m + 8.3754 N·m = T * 0.635 m
12.7536 N·m = T * 0.635 m
Now, we can solve for T (the tension in the cord):
T = 12.7536 N·m / 0.635 m
T ≈ 20.106 N
Therefore, the vertical component of the force exerted on the bar by the wall is approximately 20.106 N.