A uniform beam of length L = 8.00 m and weight 3.80 102 N is carried by two workers, Sam and Joe, as shown in the figure below. Determine the force that each person exerts on the beam.
To determine the force that each person exerts on the beam, we need to consider the weight of the beam and the distribution of the weight between the two workers.
Let's denote the force that Sam exerts on the beam as F_s and the force that Joe exerts as F_j.
Since the beam is uniform and its weight is given as 3.80 x 10^2 N, the weight is distributed evenly along its length. Therefore, each person supports half of the weight of the beam.
Using this information, we can say:
F_s + F_j = 3.80 x 10^2 N (equation 1)
We also know that the length of the beam is L = 8.00 m.
Considering the torques about the pivot point where Sam is standing, we can say:
F_j x L = F_s x (L/2) (equation 2)
By dividing equation 2 by L, we get:
F_j = F_s/2 (equation 3)
Substituting equation 3 into equation 1, we have:
F_s + F_s/2 = 3.80 x 10^2 N
Multiplying through by 2, we get:
2F_s + F_s = 760 N
Combining like terms, we have:
3F_s = 760 N
Dividing by 3, we find:
F_s = 760 N / 3 = 253.33 N
Substituting the value of F_s into equation 3, we find:
F_j = 253.33 N / 2 = 126.67 N
Therefore, Sam exerts a force of 253.33 N on the beam, while Joe exerts a force of 126.67 N on the beam.
To determine the force that each person exerts on the beam, we need to consider the forces acting on the beam and apply the principle of equilibrium.
In this case, since the beam is uniform, we can assume that its weight is acting at its center of mass, which is at the midpoint of the beam. This means that the weight of the beam is acting downward at a distance of L/2 = 8.00 m / 2 = 4.00 m from each end.
Let's denote the force exerted by Sam as FS and the force exerted by Joe as FJ. We need to find these forces.
Now, applying the principle of equilibrium, we know that the sum of the forces in the vertical direction and the sum of the torques about any point must be zero.
Considering the forces in the vertical direction, we have the following equation:
FS + FJ - weight of the beam = 0
Substituting the weight of the beam, which is given as 3.80 × 10^2 N, we get:
FS + FJ - 3.80 × 10^2 N = 0
To determine the torques, we need to choose a point about which to calculate them. Let's choose the left end of the beam as the reference point. This means that the torque due to the weight of the beam is zero at this point.
The torque due to FS is given by:
TorqueFS = FS × 0 = 0
The torque due to FJ is given by:
TorqueFJ = FJ × L/2 = FJ × 8.00 m / 2 = 4.00 × FJ
Since the sum of the torques must be zero, we have:
TorqueFS + TorqueFJ = 0
0 + 4.00 × FJ = 0
4.00 × FJ = 0
FJ = 0
Substituting this in the equation for the forces, we get:
FS + FJ - 3.80 × 10^2 N = 0
FS + 0 - 3.80 × 10^2 N = 0
FS = 3.80 × 10^2 N
Therefore, the force that Sam exerts on the beam is 3.80 × 10^2 N, and the force that Joe exerts on the beam is 0 N.