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. (beam 1 is to the left, and beam 2 is on the right.)
How much upward force does support 1 exert on the beam?
How much upward force does support 2 exert on the beam?
To find out how much upward force each support exerts on the beam, we need to consider the forces acting on the beam and the equilibrium conditions.
1. The forces acting on the beam are its weight (due to gravity) and the forces exerted by the two supports.
2. The weight of the beam is given by the formula W = m × g, where m is the mass of the beam and g is the acceleration due to gravity (approximately 9.8 m/s^2). So, the weight of the beam is W = 130 kg × 9.8 m/s^2 = 1274 N.
3. Let's consider the forces exerted by the two supports. Since the beam is in equilibrium, the sum of the forces in the vertical direction must be zero.
4. Let's assume the upward force exerted by support 1 is F1 and the upward force exerted by support 2 is F2.
5. Considering the forces in the vertical direction, we have the equation F1 + F2 - W = 0.
6. However, we can also consider the moments (torques) acting on the beam. Since the beam is not rotating, the sum of the moments must be zero.
7. The moment exerted by support 1 is given by the equation Moment1 = F1 × (distance from support 1 to the student). In this case, the distance is 2.0 m.
8. The moment exerted by support 2 is given by the equation Moment2 = F2 × (distance from support 2 to the student). In this case, the distance is (3.0 m - 2.0 m) = 1.0 m.
9. Considering the moments, we have the equation Moment1 - Moment2 = 0, which can be rewritten as F1 × 2.0 m - F2 × 1.0 m = 0.
10. Solving these equations simultaneously, we can find the values of F1 and F2:
- From the equation F1 + F2 - W = 0, we can rewrite it as F1 + F2 = W, which gives us F1 + F2 = 1274 N.
- From the equation F1 × 2.0 m - F2 × 1.0 m = 0, we can rewrite it as 2F1 - F2 = 0, which gives us F1 = (F2 × 1.0 m) / 2.0 m.
11. Substituting the value of F1 in terms of F2 in the equation F1 + F2 = 1274 N gives us (F2 × 1.0 m) / 2.0 m + F2 = 1274 N.
12. Solving this equation, we find that F2 = 847 N.
13. Substituting the value of F2 into the equation F1 = (F2 × 1.0 m) / 2.0 m gives us F1 = (847 N × 1.0 m) / 2.0 m, which simplifies to F1 = 423.5 N.
Therefore, the upward force exerted by support 1 on the beam is approximately 423.5 N, and the upward force exerted by support 2 on the beam is approximately 847 N.
To determine the upward forces exerted by support 1 and support 2 on the beam, you need to consider the equilibrium condition of the system.
First, let's calculate the weight of the beam:
Weight of the beam = mass x gravitational acceleration
Weight of the beam = 130 kg x 9.8 m/s^2 = 1274 N
Next, let's calculate the weight of the student:
Weight of the student = mass x gravitational acceleration
Weight of the student = 90 kg x 9.8 m/s^2 = 882 N
To find the upward force exerted by support 1, we need to sum up all the vertical forces acting on the beam:
ΣFy = F1 + F2 + weight of the beam + weight of the student = 0
Since the beam is in equilibrium, the sum of all vertical forces must be zero.
Substituting the known values:
F1 + F2 + 1274 N + 882 N = 0
Now, we can solve for F1, the upward force exerted by support 1:
F1 = -(F2 + 1274 N + 882 N)
To calculate the upward force exerted by support 2 on the beam, we can use the law of equilibrium:
ΣFy = F1 + F2 + weight of the beam + weight of the student = 0
Since the beam is in equilibrium, the sum of all vertical forces must be zero.
Substituting the known values:
F1 + F2 + 1274 N + 882 N = 0
Now, we can solve for F2, the upward force exerted by support 2:
F2 = -(F1 + 1274 N + 882 N)
To find the magnitudes of F1 and F2, we need to substitute the equations above and solve for the respective variables.
Note: The negative sign indicates that the forces are acting in the opposite direction to the positive direction of the y-axis.