Question
1. A 3.50-kg block on a smooth tabletop is attached by a string to a hanging block of mass
2.80 kg, as shown. The blocks are released from rest and allowed to move freely. (a) Is
the tension in the string greater than, less than, or equal to the weight of the hanging
mass? Find (b) the acceleration of the blocks and (c) the tension in the string.
2. To hang a 6.20-kg pot of flowers, a gardener uses two wires—one attached horizontally
to a wall, the other sloping upward at an angle of and attached to the ceiling.
Find the tension in each wire.
3. A traction device employing three pulleys is applied to a broken leg, as shown in the
sketch. The middle pulley is attached to the sole of the foot, and a mass m supplies the
tension in the ropes. Find the value of the mass m if the force exerted on the sole of the
foot by the middle pulley is to be 165 N
2.80 kg, as shown. The blocks are released from rest and allowed to move freely. (a) Is
the tension in the string greater than, less than, or equal to the weight of the hanging
mass? Find (b) the acceleration of the blocks and (c) the tension in the string.
2. To hang a 6.20-kg pot of flowers, a gardener uses two wires—one attached horizontally
to a wall, the other sloping upward at an angle of and attached to the ceiling.
Find the tension in each wire.
3. A traction device employing three pulleys is applied to a broken leg, as shown in the
sketch. The middle pulley is attached to the sole of the foot, and a mass m supplies the
tension in the ropes. Find the value of the mass m if the force exerted on the sole of the
foot by the middle pulley is to be 165 N
Answers
GPT-4o mini
Let's tackle each problem one by one.
### Problem 1:
We have two blocks: Block \( A \) with mass \( m_A = 3.50 \) kg on a tabletop and Block \( B \) with mass \( m_B = 2.80 \) kg hanging vertically.
#### (a) Tension Analysis
1. The forces acting on mass \( B \) (hanging mass) are:
- Weight: \( F_g = m_B \cdot g = 2.80 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 27.548 \, \text{N} \)
- Tension: \( T \)
From the perspective of mass \( B \), we have:
\[
m_B g - T = m_B a \text{ (Equation for mass B)}
\]
2. The forces acting on mass \( A \) (on the tabletop) are:
- Tension: \( T \)
- The only force acting (assuming no friction): \( T = m_A a \)
From the perspective of mass \( A \), we have:
\[
T = m_A a \text{ (Equation for mass A)}
\]
**Comparison of Tension with Weight**:
- Since \( m_B g \) is greater than the tension (which is responsible for acceleration), the tension \( T \) is **less than the weight of the hanging mass**.
#### (b) Acceleration of the Blocks
To find \( a \), we'll substitute \( T \) from the second equation into the first equation.
1. Substitute \( T = m_A a \) into \( m_B g - T = m_B a \):
\[
m_B g - m_A a = m_B a
\]
2. Rearranging gives:
\[
m_B g = m_B a + m_A a
\]
\[
m_B g = (m_A + m_B) a
\]
\[
a = \frac{m_B g}{m_A + m_B}
\]
\[
a = \frac{(2.80 \, \text{kg}) (9.81 \, \text{m/s}^2)}{3.50 \, \text{kg} + 2.80 \, \text{kg}} = \frac{27.548}{6.30} \approx 4.375 \, \text{m/s}^2
\]
#### (c) Tension in the String
Now, use the acceleration to find \( T \):
\[
T = m_A a = 3.50 \cdot 4.375 \approx 15.31 \, \text{N}
\]
### Summary for Problem 1:
- (a) Tension \( T < m_B g \)
- (b) \( a \approx 4.38 \, \text{m/s}^2 \)
- (c) \( T \approx 15.31 \, \text{N} \)
---
### Problem 2:
We need to calculate the tensions in two wires supporting a pot of flowers weighing \( 6.20 \, \text{kg} \).
Assuming the angle of the sloping wire relative to the horizontal is \( \theta \).
1. The weight of the pot is:
\[
F_g = m \cdot g = 6.20 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 60.82 \, \text{N}
\]
2. If \( T_1 \) is the tension in the horizontal wire and \( T_2 \) is the tension in the sloping wire:
\[
T_2 \sin \theta = F_g \quad \text{(Vertical equilibrium)}
\]
\[
T_1 = T_2 \cos \theta \quad \text{(Horizontal equilibrium)}
\]
You will need to provide the angle \( \theta \) in order to solve for \( T_1 \) and \( T_2 \).
---
### Problem 3:
For this traction device with three pulleys and a force \( F = 165 \, \text{N} \):
1. Let \( m \) be the mass providing the tension in the ropes. The total force exerted on the sole of the foot due to pulleys is:
\[
F = m \cdot g \cdot \text{(Number of sections of rope)}
\]
If there are \( n = 3 \) sections of rope supporting the mass:
\[
165 \, \text{N} = m \cdot g / 3
\]
So,
\[
m \cdot g = 3 \cdot 165 \, \text{N}
\]
\[
m \cdot 9.81 \, \text{m/s}^2 = 495 \, \text{N}
\]
\[
m = \frac{495}{9.81} \approx 50.43 \, \text{kg}
\]
### Summary for Problem 3:
- The mass \( m \approx 50.43 \, \text{kg} \).
If you have specific numbers, angles, or conditions for problems 2 and 3, please provide them!
### Problem 1:
We have two blocks: Block \( A \) with mass \( m_A = 3.50 \) kg on a tabletop and Block \( B \) with mass \( m_B = 2.80 \) kg hanging vertically.
#### (a) Tension Analysis
1. The forces acting on mass \( B \) (hanging mass) are:
- Weight: \( F_g = m_B \cdot g = 2.80 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 27.548 \, \text{N} \)
- Tension: \( T \)
From the perspective of mass \( B \), we have:
\[
m_B g - T = m_B a \text{ (Equation for mass B)}
\]
2. The forces acting on mass \( A \) (on the tabletop) are:
- Tension: \( T \)
- The only force acting (assuming no friction): \( T = m_A a \)
From the perspective of mass \( A \), we have:
\[
T = m_A a \text{ (Equation for mass A)}
\]
**Comparison of Tension with Weight**:
- Since \( m_B g \) is greater than the tension (which is responsible for acceleration), the tension \( T \) is **less than the weight of the hanging mass**.
#### (b) Acceleration of the Blocks
To find \( a \), we'll substitute \( T \) from the second equation into the first equation.
1. Substitute \( T = m_A a \) into \( m_B g - T = m_B a \):
\[
m_B g - m_A a = m_B a
\]
2. Rearranging gives:
\[
m_B g = m_B a + m_A a
\]
\[
m_B g = (m_A + m_B) a
\]
\[
a = \frac{m_B g}{m_A + m_B}
\]
\[
a = \frac{(2.80 \, \text{kg}) (9.81 \, \text{m/s}^2)}{3.50 \, \text{kg} + 2.80 \, \text{kg}} = \frac{27.548}{6.30} \approx 4.375 \, \text{m/s}^2
\]
#### (c) Tension in the String
Now, use the acceleration to find \( T \):
\[
T = m_A a = 3.50 \cdot 4.375 \approx 15.31 \, \text{N}
\]
### Summary for Problem 1:
- (a) Tension \( T < m_B g \)
- (b) \( a \approx 4.38 \, \text{m/s}^2 \)
- (c) \( T \approx 15.31 \, \text{N} \)
---
### Problem 2:
We need to calculate the tensions in two wires supporting a pot of flowers weighing \( 6.20 \, \text{kg} \).
Assuming the angle of the sloping wire relative to the horizontal is \( \theta \).
1. The weight of the pot is:
\[
F_g = m \cdot g = 6.20 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 60.82 \, \text{N}
\]
2. If \( T_1 \) is the tension in the horizontal wire and \( T_2 \) is the tension in the sloping wire:
\[
T_2 \sin \theta = F_g \quad \text{(Vertical equilibrium)}
\]
\[
T_1 = T_2 \cos \theta \quad \text{(Horizontal equilibrium)}
\]
You will need to provide the angle \( \theta \) in order to solve for \( T_1 \) and \( T_2 \).
---
### Problem 3:
For this traction device with three pulleys and a force \( F = 165 \, \text{N} \):
1. Let \( m \) be the mass providing the tension in the ropes. The total force exerted on the sole of the foot due to pulleys is:
\[
F = m \cdot g \cdot \text{(Number of sections of rope)}
\]
If there are \( n = 3 \) sections of rope supporting the mass:
\[
165 \, \text{N} = m \cdot g / 3
\]
So,
\[
m \cdot g = 3 \cdot 165 \, \text{N}
\]
\[
m \cdot 9.81 \, \text{m/s}^2 = 495 \, \text{N}
\]
\[
m = \frac{495}{9.81} \approx 50.43 \, \text{kg}
\]
### Summary for Problem 3:
- The mass \( m \approx 50.43 \, \text{kg} \).
If you have specific numbers, angles, or conditions for problems 2 and 3, please provide them!