The 3.0 m long, 94 kg rigid beam in the following figure is supported at each end. An 70 kg student stands 2.0 m from support 1.
How much upward force does the support 1 exert on the beam?
How much upward force does the support 2 exert on the beam?
To determine the upward force exerted by each support on the beam, we need to consider the forces acting on the beam and the equilibrium condition.
Let's denote the upward force exerted by support 1 as F1 and the upward force exerted by support 2 as F2.
Step 1: Analyze the forces acting on the beam
- The weight of the beam acts downward and can be calculated using the formula: weight = mass × gravitational acceleration (g).
Given the mass of the beam as 94 kg, and the acceleration due to gravity as approximately 9.8 m/s^2, the weight of the beam can be calculated as: weight = 94 kg × 9.8 m/s^2.
- The weight of the student also acts downward and can be calculated using the same formula: weight = mass × gravitational acceleration (g).
Given the mass of the student as 70 kg, and the acceleration due to gravity as approximately 9.8 m/s^2, the weight of the student can be calculated as: weight = 70 kg × 9.8 m/s^2.
- The beam exerts two reaction forces, F1 and F2, on the supports. These forces act upward to balance the downward forces acting on the beam.
Step 2: Apply equilibrium condition
For the beam to be in equilibrium, the sum of the forces acting on it in both the vertical and horizontal directions must be zero.
In the vertical direction, we have:
Sum of upward forces = Sum of downward forces.
- Upward forces: F1 + F2
- Downward forces: weight of beam + weight of student
Step 3: Calculate the forces
Using the equilibrium condition, we can now determine the values of F1 and F2.
Based on the information provided, the student stands 2.0 m from support 1. Since the beam is symmetric, the distance between the student and support 2 is also 2.0 m.
To calculate the weight of the beam and student, use the given masses and gravitational acceleration:
- weight of beam = 94 kg × 9.8 m/s^2
- weight of student = 70 kg × 9.8 m/s^2
Finally, solve the equilibrium equation to determine the values of F1 and F2 by substituting the known values:
F1 + F2 = (weight of beam) + (weight of student)
Substitute the calculated values of the weights and solve for F1 and F2.
To find the upward force exerted by each support on the beam, we need to analyze the forces acting on the beam and apply the principle of torque equilibrium.
1. First, let's consider the forces acting on the beam. These include:
- The weight of the beam, acting downward at its center.
- The weight of the student, acting downward at the student's location.
- The upward forces exerted by the supports.
2. Since the beam is in equilibrium, the sum of all the torques acting on it must be zero. To determine the torques, we need to choose a pivot point. The pivot point should be somewhere on the beam where it is known that there will be no torque contribution from the supports.
3. Let's choose the pivot point at Support 2. This means that the torque due to Support 2 will be zero, and we can focus on the torque due to Support 1 and the weights.
4. The torque (T) acting on an object is calculated by multiplying the force (F) applied perpendicular to the lever arm (r): T = F * r. The lever arm is the perpendicular distance between the force and the pivot point.
5. The torque due to Support 1 is zero because the line of action of its force passes through the pivot point. Therefore, Support 1 exerts no torque on the beam.
6. Now, let's calculate the torque due to the weight of the beam. The weight of the beam acts at its center, which is at a distance of half its length (1.5 m) from either support.
Torque due to the weight of the beam = Weight of the beam * Distance to the pivot point
= (94 kg * 9.8 m/s^2) * (1.5 m).
7. Next, let's calculate the torque due to the weight of the student. The weight of the student acts at a distance of 2.0 m from Support 1.
Torque due to the weight of the student = Weight of the student * Distance to the pivot point
= (70 kg * 9.8 m/s^2) * (2.0 m).
8. Since the beam is in equilibrium, the sum of the torques must be zero. Therefore, we can write the equation:
Torque due to the weight of the beam + Torque due to the weight of the student = 0.
9. By substituting the values, we can solve for the upward force exerted by Support 1 on the beam. However, to simplify the calculation, the downward forces can be treated as negative values, and the upward forces can be treated as positive values.
F_support1 * 3.0 m - (94 kg * 9.8 m/s^2) * 1.5 m - (70 kg * 9.8 m/s^2) * 2.0 m = 0.
Solving this equation will give you the magnitude of the upward force exerted by Support 1 on the beam.
10. To find the upward force exerted by Support 2 on the beam, simply subtract the upward force exerted by Support 1 from the total weight of the beam and the student.
F_support2 = (94 kg * 9.8 m/s^2) + (70 kg * 9.8 m/s^2) - F_support1.
This will give you the magnitude of the upward force exerted by Support 2 on the beam.
By following these steps and calculating the values, you can find the upward force exerted by Support 1 and Support 2 on the beam.
alternately use support 1 and support 2 as a fulcrum
calculate the moments
the mass of the beam acts in the center