A 3.0-m-long rigid beam with a mass 130 kg is supported at each end. An 90 kg student stands 2.0 m from support 1.
How much upward force does support 1 exert on the beam?
How much upward force does support 2 exert on the beam?
Well, let's see. If the 90 kg student is standing 2.0 m from support 1, I hope they're not putting too much weight on that poor beam! But let's do the math anyway.
To find the upward force exerted by support 1, we need to consider the forces acting on the beam. We have the weight of the beam itself, which is given as 130 kg, and the weight of the student, which is 90 kg.
Now, since the beam is being supported at both ends, the supports are sharing the load. So, we can assume that the weight of the beam is distributed evenly between the two supports.
Therefore, the weight supported by support 1 would be half of the total weight of the beam plus the weight of the student. Can you guess how much that is? Well, it's 130 kg divided by 2, plus 90 kg.
As for the second question, since the weight is being shared equally between the supports, the upward force exerted by support 2 would also be the same as support 1.
So, in conclusion, both support 1 and support 2 would exert the same upward force, which is equal to half the weight of the beam plus the weight of the student.
To determine the upward force exerted by each support on the beam, we need to consider the equilibrium of forces.
1. First, let's find the weight of the beam. The weight is given by the formula:
Weight = mass * gravitational acceleration
Given:
Mass of the beam = 130 kg
Gravitational acceleration = 9.8 m/s^2
Weight of the beam = 130 kg * 9.8 m/s^2 = 1274 N
2. Next, let's find the weight of the student. Again, using the formula:
Weight = mass * gravitational acceleration
Given:
Mass of the student = 90 kg
Gravitational acceleration = 9.8 m/s^2
Weight of the student = 90 kg * 9.8 m/s^2 = 882 N
3. Now, let's calculate the forces exerted by each support on the beam.
On one side of the beam, the weight of the beam and the weight of the student acting on it create a clockwise moment (due to their distance from support 1), which needs to be balanced by an anticlockwise moment created by the upward force from support 1.
Let's denote the upward force exerted by support 1 as F1.
Using the moment equation:
Clockwise moment = Anticlockwise moment
(Weight of the beam * Distance of the beam's center of mass from support 1) + (Weight of the student * Distance of the student from support 2) = F1 * Length of the beam
(1274 N * (3 m / 2)) + (882 N * (2 m)) = F1 * 3 m
(1903 N) + (1764 N) = F1 * 3 m
3667 N = F1 * 3 m
F1 = 3667 N / 3 m
F1 = 1222.33 N
Therefore, the upward force exerted by support 1 on the beam is approximately 1222.33 N.
4. To find the upward force exerted by support 2, we need to consider the overall equilibrium of forces on the beam.
Since the two supports are symmetrically placed, the upward force exerted by support 2 will be the same as that exerted by support 1.
Therefore, the upward force exerted by support 2 on the beam is also approximately 1222.33 N.
To find the upward forces exerted by each support on the beam, we need to consider the torque equilibrium.
1. To determine the upward force exerted by support 1 on the beam, we need to calculate the torque.
Torque is the product of force and the perpendicular distance from the point of rotation. In this case, the point of rotation is support 2.
Let's call the distance from support 1 to support 2 as L, which is 3.0 m in this case.
The torque exerted by the student is the product of the student's weight (mg) and the distance from support 2, which is 3.0 m - 2.0 m = 1.0 m.
Hence, the torque exerted by the student is (90 kg * 9.8 m/s^2) * 1.0 m = 882 N*m.
To maintain equilibrium, the total torque exerted by the beam and the support forces must be zero. Since support 1 is exerting an upward force, the torque it exerts is in the clockwise direction. This means the torque exerted by support 2 is in the counterclockwise direction.
The torque exerted by support 2 is (Force2) * L. Since it is in equilibrium, the torque exerted by support 2 is equal to the torque exerted by the student.
Therefore, (Force2) * L = 882 N*m.
2. To determine the upward force exerted by support 1, we can apply the principle of equilibrium.
The sum of the vertical forces acting on the beam must be zero.
Considering upward forces as positive and downward forces as negative, the equation becomes:
(Force1) + (-90 kg * 9.8 m/s^2) = 0.
Simplifying, we get:
(Force1) = 90 kg * 9.8 m/s^2 = 882 N.
Hence, support 1 exerts an upward force of 882 N on the beam.
3. To determine the upward force exerted by support 2, we can rearrange the torque equation we derived earlier:
(Force2) * L = 882 N*m.
Since L is 3.0 m, we can solve for Force2:
(Force2) = 882 N*m / 3.0 m.
(Force2) = 294 N.
Therefore, support 2 exerts an upward force of 294 N on the beam.
Becasue vertical frce must balance,
F1 + F2 = (M1 + M2)* g
M1 is the beam mass. M2 is the student's mass.
F1 is the force on support 1.
F2 is the force on support 2.
Now write a moment balance equation, such as.
M2*g*2 + M1*g*1.5 = F2*3
You can solve the last equation for F2 directly. Then use the first equation to solve for F1.