Question
A)A lift of 950kg is carrying a woman of mass 50kg.
The lift is ascending at a uniform speed. Calculate:
Tension in the lift cable
Vertical force exerted on the woman by the floor of the lift
B) Sometimes later, the lift is ascending with a downward acceleration of 2ms^-2. Calculate the tension in the lift cable.
Vertical force exerted on the woman by the lift
The lift is ascending at a uniform speed. Calculate:
Tension in the lift cable
Vertical force exerted on the woman by the floor of the lift
B) Sometimes later, the lift is ascending with a downward acceleration of 2ms^-2. Calculate the tension in the lift cable.
Vertical force exerted on the woman by the lift
Answers
GPT-4o mini
To solve the problems, we will use the principles of mechanics, particularly Newton's second law, which states that \( F = ma \).
### A) The lift is ascending at a uniform speed
1. **Tension in the lift cable:**
When the lift is moving at a uniform speed, the net force acting on it is zero (i.e., the forces are balanced). The forces acting on the lift are its weight and the tension in the cable.
The total mass being lifted by the cable is:
\[
m_{\text{total}} = m_{\text{lift}} + m_{\text{woman}} = 950 \, \text{kg} + 50 \, \text{kg} = 1000 \, \text{kg}
\]
The weight of the lift plus the woman is:
\[
W = m_{\text{total}} \cdot g = 1000 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 9810 \, \text{N}
\]
Since the lift is ascending at a uniform speed (constant velocity), the tension \( T \) in the cable is equal to the weight:
\[
T = W = 9810 \, \text{N}
\]
2. **Vertical force exerted on the woman by the floor of the lift:**
The only vertical forces acting on the woman are her weight and the normal force (the force exerted by the lift floor). Since the woman is also at a constant velocity (not accelerating), the forces are balanced as well.
The weight of the woman is:
\[
W_{\text{woman}} = m_{\text{woman}} \cdot g = 50 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 490.5 \, \text{N}
\]
Therefore, the normal force (vertical force exerted on the woman) must also equal her weight:
\[
N = W_{\text{woman}} = 490.5 \, \text{N}
\]
### B) The lift is ascending with a downward acceleration of \( 2 \, \text{ms}^{-2} \)
1. **Tension in the lift cable:**
When the lift is accelerating downwards, the net force acting on the lift must be taken into account. The net force can be expressed as:
\[
F_{\text{net}} = m_{\text{total}} \cdot a
\]
where \( a = 2 \, \text{ms}^{-2} \). The net downward force on the lift is:
\[
F_{\text{net}} = 1000 \, \text{kg} \cdot 2 \, \text{ms}^{-2} = 2000 \, \text{N}
\]
The weight of the lift and the woman (as calculated earlier) is \( 9810 \, \text{N} \).
By using Newton's second law:
\[
T + F_{\text{net}} = W
\]
Rearranging, we find:
\[
T = W - F_{\text{net}} = 9810 \, \text{N} - 2000 \, \text{N} = 7810 \, \text{N}
\]
2. **Vertical force exerted on the woman by the lift:**
The vertical forces acting on the woman also need to be balanced. The normal force \( N \) (the force exerted by the lift floor on the woman) now acts upward against both her weight and the effect of the lift's downward acceleration.
The effective weight on the woman due to the downward acceleration of the lift is:
\[
W' = m_{\text{woman}} \cdot (g - a) = 50 \, \text{kg} \cdot (9.81 \, \text{m/s}^2 - 2 \, \text{m/s}^2) = 50 \, \text{kg} \cdot 7.81 \, \text{m/s}^2 = 390.5 \, \text{N}
\]
Thus, the normal force equals the effective weight of the woman:
\[
N = W' = 390.5 \, N
\]
### Summary of Results:
- **For part A:**
- Tension in the lift cable: \( 9810 \, \text{N} \)
- Vertical force on the woman: \( 490.5 \, \text{N} \)
- **For part B:**
- Tension in the lift cable: \( 7810 \, \text{N} \)
- Vertical force on the woman: \( 390.5 \, \text{N} \)
### A) The lift is ascending at a uniform speed
1. **Tension in the lift cable:**
When the lift is moving at a uniform speed, the net force acting on it is zero (i.e., the forces are balanced). The forces acting on the lift are its weight and the tension in the cable.
The total mass being lifted by the cable is:
\[
m_{\text{total}} = m_{\text{lift}} + m_{\text{woman}} = 950 \, \text{kg} + 50 \, \text{kg} = 1000 \, \text{kg}
\]
The weight of the lift plus the woman is:
\[
W = m_{\text{total}} \cdot g = 1000 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 9810 \, \text{N}
\]
Since the lift is ascending at a uniform speed (constant velocity), the tension \( T \) in the cable is equal to the weight:
\[
T = W = 9810 \, \text{N}
\]
2. **Vertical force exerted on the woman by the floor of the lift:**
The only vertical forces acting on the woman are her weight and the normal force (the force exerted by the lift floor). Since the woman is also at a constant velocity (not accelerating), the forces are balanced as well.
The weight of the woman is:
\[
W_{\text{woman}} = m_{\text{woman}} \cdot g = 50 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 490.5 \, \text{N}
\]
Therefore, the normal force (vertical force exerted on the woman) must also equal her weight:
\[
N = W_{\text{woman}} = 490.5 \, \text{N}
\]
### B) The lift is ascending with a downward acceleration of \( 2 \, \text{ms}^{-2} \)
1. **Tension in the lift cable:**
When the lift is accelerating downwards, the net force acting on the lift must be taken into account. The net force can be expressed as:
\[
F_{\text{net}} = m_{\text{total}} \cdot a
\]
where \( a = 2 \, \text{ms}^{-2} \). The net downward force on the lift is:
\[
F_{\text{net}} = 1000 \, \text{kg} \cdot 2 \, \text{ms}^{-2} = 2000 \, \text{N}
\]
The weight of the lift and the woman (as calculated earlier) is \( 9810 \, \text{N} \).
By using Newton's second law:
\[
T + F_{\text{net}} = W
\]
Rearranging, we find:
\[
T = W - F_{\text{net}} = 9810 \, \text{N} - 2000 \, \text{N} = 7810 \, \text{N}
\]
2. **Vertical force exerted on the woman by the lift:**
The vertical forces acting on the woman also need to be balanced. The normal force \( N \) (the force exerted by the lift floor on the woman) now acts upward against both her weight and the effect of the lift's downward acceleration.
The effective weight on the woman due to the downward acceleration of the lift is:
\[
W' = m_{\text{woman}} \cdot (g - a) = 50 \, \text{kg} \cdot (9.81 \, \text{m/s}^2 - 2 \, \text{m/s}^2) = 50 \, \text{kg} \cdot 7.81 \, \text{m/s}^2 = 390.5 \, \text{N}
\]
Thus, the normal force equals the effective weight of the woman:
\[
N = W' = 390.5 \, N
\]
### Summary of Results:
- **For part A:**
- Tension in the lift cable: \( 9810 \, \text{N} \)
- Vertical force on the woman: \( 490.5 \, \text{N} \)
- **For part B:**
- Tension in the lift cable: \( 7810 \, \text{N} \)
- Vertical force on the woman: \( 390.5 \, \text{N} \)