A pulley is attached to the ceiling (as shown in the figure below). Spring scale A is attached to the wall and a rope runs horizontally from it and over the pulley. The same rope is then attached to spring scale B. On the other side of scale B hangs an W = 75-N weight. What are the readings of the two scales A and B? The weights of the scales are negligible.
75 N
To solve this problem, we need to consider the equilibrium conditions of the system. In this case, the sum of the forces acting on each scale must be zero.
Let's start by analyzing scale A:
1. The tension in the rope attached to scale A pulls it horizontally towards the left.
2. As the system is in equilibrium, this tension must be balanced by an equal and opposite force acting on scale A from the wall.
3. Therefore, the reading on scale A is equal to the tension in the rope.
Next, let's analyze scale B:
1. Since the rope is attached to both scale A and scale B, the tension in the rope is the same throughout its length.
2. The rope is connected to scale B vertically, and the weight W hangs down from it. The tension in the rope must support the weight of W.
3. Therefore, the reading on scale B is equal to the weight of W, which is 75 N.
In conclusion, the reading on scale A is equal to the tension in the rope, and the reading on scale B is equal to the weight of the object hanging from it, which is 75 N.
To find the readings of the two scales, we need to analyze the forces acting on the system.
Let's assign variables to the unknowns:
- Reading on spring scale A: A
- Reading on spring scale B: B
Based on the problem description, we know that there is a weight W = 75 N hanging from the end of spring scale B.
Now, let's break down the forces acting on the system:
1. Weight W (75 N): This force acts downwards, pulling spring scale B downwards.
2. Tension in the rope: The tension in the rope acts in both directions and is the same throughout its length. The tension pulls spring scale A towards the right and spring scale B towards the left.
3. Support force on the ceiling: The ceiling provides a support force to balance the tensions in the rope. This force acts upwards and cancels out the vertical component of the tensions.
Now, let's analyze the tension in the rope:
- Since the rope is inextensible and connected to scale A and scale B, the tension in the rope must be the same at both ends. Let's call this tension T.
Now, let's consider the forces acting on scale A:
- The only force acting on scale A is the tension T directed to the right.
Resultant force on scale A: F_A = T
Now, let's consider the forces acting on scale B:
- The weight W acts downwards, creating a force equal to 75 N.
- The tension T in the rope acts towards the left.
Resultant force on scale B: F_B = T - W
In equilibrium, the resultant force on both scales should be zero:
- F_A = 0
- F_B = 0
Since the forces are balanced, we can set up the following equations:
- F_A = T = 0
- F_B = T - W = 0
Now, let's substitute the value of W = 75 N into the equation for F_B:
- T - 75 = 0
We can now solve for T:
- T = 75 N
Finally, we can calculate the readings on the scales:
- Reading on scale A: A = T = 75 N
- Reading on scale B: B = T + W = 75 N + 75 N = 150 N
Therefore, the reading on spring scale A is 75 N and the reading on spring scale B is 150 N.