Question

A uniform 1200 N piece of medical apparatus that is 3.5 m long is suspended horizontally by two vertical wires at its ends. A small but dense 550 N weight is placed on the apparatus 2.0 m from one end, as shown in the figure. What are the tensions, A and B, in the two wires?

Answer 1

To find the tensions in the two wires suspending the medical apparatus, we can solve for the forces acting on the apparatus. Using the concept of torque, we can determine that the tensions in the two wires are equal.

Step-by-step explanation:

To find the tensions in the two wires, we can solve for the forces acting at the ends of the apparatus. Let's call the tension in wire A TA and the tension in wire B TB. From the given information, we know that the weight of the apparatus is 1200 N and it is suspended horizontally. The weight of the small weight is 550 N, and it is placed 2.0 m from one end of the apparatus.

Since the apparatus is in equilibrium, the sum of the horizontal forces acting on it must be zero. This means that the tension in wire A is equal to the tension in wire B.

Using the concept of torque, we can determine the tensions in the wires. The torque due to the weight of the apparatus must be balanced by the torque due to the weight of the small weight. The torque due to the weight of the apparatus is given by the equation:

Torque = force × perpendicular distance from the axis of rotation

Since the apparatus is suspended horizontally, the distance from the axis of rotation to the weight of the apparatus is half of its length, which is 1.75 m. The torque due to the weight of the apparatus is therefore:

Torqueapparatus = 1200 N × 1.75 m

The torque due to the weight of the small weight is:

Torque = 550 N × 2.0 m

Since the torque due to the weight of the apparatus must be balanced by the torque due to the weight of the small weight, we have:

Torqueapparatus = Torque

1200 N × 1.75 m = 550 N × 2.0 m

Simplifying this equation, we find:

2100 Nm = 1100 Nm

Since the torque on both sides is equal, the tensions in the wires must also be equal. Therefore, we have:

TA = TB

Therefore, the tensions in the two wires are equal.

Answer 2

To determine the tensions in the two wires suspending the medical apparatus, we can analyze the forces acting on it.

Let's consider the apparatus to be divided into two sections: Section A (with length 2.0 m) and Section B (with length 1.5 m, which is the remaining length).

In Section A:
- The weight of the apparatus (1200 N) exerts a downward force at the center of gravity, which is 1.0 m from each end of Section A.
- The small weight (550 N) is placed 2.0 m from one end, which means it exerts a downward force 2.0 m from the same end.

In Section B:
- Only the weight of the apparatus (1200 N) exerts a downward force at the center of gravity, which is at the midpoint of Section B (0.75 m from each end).

Considering equilibrium, the sum of the clockwise moments (due to the forces) should be equal to the sum of the counterclockwise moments.

Let's calculate the tensions in the wires. Let T_A represent the tension in wire A and T_B represent the tension in wire B.

Clockwise moments (due to the forces):
- Weight of the apparatus in Section A: 1200 N * (2.0 m) = 2400 N·m
- Small weight in Section A: 550 N * (2.0 m) = 1100 N·m

Counterclockwise moments:
- Weight of the apparatus in Section B: 1200 N * (0.75 m) = 900 N·m

For equilibrium, the clockwise moments should be equal to the counterclockwise moments:
2400 N·m + 1100 N·m = 900 N·m

To find the tensions in the wires, we can consider the vertical forces:
In the vertical direction:
T_A + T_B = weight of the apparatus (1200 N) + small weight (550 N) = 1750 N

Now we have two equations:
T_A + T_B = 1750 N
T_B = 1750 N - T_A

Substituting the value of T_B in the clockwise moments equation:
2400 N·m + 1100 N·m = 900 N·m + T_A * (3.5 m)

Simplifying the equation:
3500 N·m = 900 N·m + T_A * (3.5 m)

Rearranging the equation to solve for T_A:
T_A * (3.5 m) = 3500 N·m - 900 N·m
T_A * (3.5 m) = 2600 N·m
T_A = 2600 N·m / (3.5 m)
T_A ≈ 742.86 N

Now, substituting the value of T_A in the equation T_A + T_B = 1750 N:
742.86 N + T_B = 1750 N
T_B = 1750 N - 742.86 N
T_B ≈ 1007.14 N

Therefore, the tension in wire A (T_A) is approximately 742.86 N, and the tension in wire B (T_B) is approximately 1007.14 N.