A uniform beam weighing Wa=1200 N is 3.50 m long and is suspended horizontally by two vertical wires at its ends, as shown in the figure below. A small but dense weight with WL = 550 N is placed on the beam 2.00m from one end, as shown in the figure.

a) are the tensions Ta (in cable A) and TB (in cable B) equal?

b) Considering that this beam is in total equilibrium, now write the equation in rotational equilibrium.

c) In the summation below, write the magnitudes of the CW and the CCW around a point on the left end of the beam, i.e., on the side labeled as needed.

a) No they are not equal, becasue the addede load is not in the middle of the beam.

b) 1200*1.75 + 550*2 = TB*3.50
(assuming the added weight is 2m from end A) Solve that for TB

c) There is no "summation below". Also, the side where you want to compute the wire tension is NOT the point that you compute moments about. It is the opposite end.

835.7 and 914.3

a) The tensions Ta (in cable A) and TB (in cable B) are not equal. The tension in each cable will depend on the distribution of weights on the beam.

b) To write the equation for rotational equilibrium, we need to consider the moments about a chosen point. Let's choose the left end of the beam as the point. The equation for rotational equilibrium can be written as:

Στ = 0

Where Στ is the sum of the torques (or moments) acting on the beam.

c) To write the magnitudes of the clockwise (CW) and counterclockwise (CCW) torques around the left end of the beam, we need to consider the weights and their distances from the chosen point.

- The weight of the beam itself (Wa) will create a counterclockwise torque since it is acting downwards and its distance from the left end is 0. Therefore, the CCW torque caused by the beam is 0 N·m.

- The weight placed on the beam (WL) will create both CW and CCW torques. Since this weight is 2.00 m from the left end, its CCW torque can be calculated as CCW = WL * distance = 550 N * 2.00 m = 1100 N·m.

Since there are no other weights or forces mentioned in the problem, the sum of the CW and CCW torques will be equal to the CCW torque caused by the weight on the beam:

Στ = CW - CCW = 0 - 1100 N·m = -1100 N·m

a) To determine whether the tensions Ta and Tb are equal, we need to analyze the forces acting on the beam. In this case, the beam is in equilibrium, meaning that the net torque acting on it is zero.

b) To write the equation for rotational equilibrium, we can use the condition that the sum of the torques acting on the beam must be zero. The torque is calculated by multiplying the force by the lever arm (distance from the point of rotation to the point of application of the force).

c) To write the summation of the clockwise (CW) and counterclockwise (CCW) torques around a point on the left end of the beam, we need to consider the forces acting on the beam and their respective lever arms.

To calculate the tension in cable A (Ta) and cable B (Tb), we can use the principle of moments.

Let's follow these steps to answer the questions:

a) To determine if Ta and Tb are equal, we need to consider that the beam is in equilibrium. Since the beam is horizontally suspended by two vertical wires, the vertical components of the tensions in the cables (Ta and Tb) cancel each other out due to symmetry. Therefore, the remaining horizontal components of the tensions, which cause rotation, must also be equal in magnitude to maintain equilibrium. Therefore, Ta = Tb.

b) To write the equation for rotational equilibrium, we need to consider the torques acting on the beam. The torque due to the weight Wl can be calculated by multiplying its magnitude by its lever arm, which is the distance between the weight and the point of rotation (one end of the beam). The torque due to the weight is CCW (counterclockwise) because it acts in the opposite direction of a clockwise rotation.

c) To write the summation of the clockwise (CW) and counterclockwise (CCW) around a point on the left end of the beam, we need to consider the torques acting on the beam. The clockwise torques can arise from the weight of the beam itself (Wa) and the tension in cable B (Tb). The counterclockwise torque can arise from the weight placed on the beam (Wl) and the tension in cable A (Ta).

Therefore, the summation of the clockwise (CW) torques around the left end of the beam can be written as:
CW = Tb * L (torque due to tension in cable B)
CW = 0 (there is no torque due to the weight of the beam itself)

The counterclockwise torque (CCW) around the left end of the beam can be written as:
CCW = Wl * d (torque due to the weight placed on the beam)
CCW = Wa * (L/2) (torque due to the weight of the beam itself)
CCW = Ta * L (torque due to tension in cable A)

The magnitudes of the CW and CCW torques around the left end of the beam are:

CW = Tb * L
CCW = Wl * d + Wa * (L/2) + Ta * L