34.2.3 a man standing on a scale in an elevator (man and scale) 34.3.1 What determines the apparent weight of a person? $$ \begin{array}{l}34.3 .2 ext { What is the apparent weight of a man, mass } \ 60 \mathrm{~kg} ext {, accelerating downwards at } 2 \mathrm{~m} \cdot \mathrm{~s}^{-2} ext { in } \ ext { an elevator? } \ \begin{array}{l} ext { 34.3.3 When does a person experience a feeling } \ ext { of weightlessness? }\end{array} \ ext { Question } 35 \ ext { Two blocks of masses } 12 \mathrm{~kg} ext { and } 20 \mathrm{~kg} ext { are at rest } \ ext { against each other on a frictionless surface. A force } \ ext { of } 40 \mathrm{~N} ext { is applied to the larger block. }\end{array} $$

Question

34.2.3 a man standing on a scale in an elevator (man and scale) 34.3.1 What determines the apparent weight of a person? $$ \begin{array}{l}34.3 .2 	ext { What is the apparent weight of a man, mass } \ 60 \mathrm{~kg} 	ext {, accelerating downwards at } 2 \mathrm{~m} \cdot \mathrm{~s}^{-2} 	ext { in } \ 	ext { an elevator? } \ \begin{array}{l}	ext { 34.3.3 When does a person experience a feeling } \ 	ext { of weightlessness? }\end{array} \ 	ext { Question } 35 \ 	ext { Two blocks of masses } 12 \mathrm{~kg} 	ext { and } 20 \mathrm{~kg} 	ext { are at rest } \ 	ext { against each other on a frictionless surface. A force } \ 	ext { of } 40 \mathrm{~N} 	ext { is applied to the larger block. }\end{array} $$

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**34.2.3 – Man Standing on a Scale in an Elevator**  
The reading on the scale is the normal force ($$N$$) that the scale exerts on the man. When the elevator is accelerating, the normal force is given by  
$$
N = mg - ma
$$  
if the elevator accelerates downward (or  
$$
N = mg + ma
$$  
if it accelerates upward), where:  
- $$m$$ is the mass of the man,  
- $$g$$ is the acceleration due to gravity, and  
- $$a$$ is the magnitude of the acceleration of the elevator.

---

**34.3.1 – What Determines the Apparent Weight of a Person?**  
The apparent weight is determined by the normal force acting on the person. In other words, it depends on how much force is needed to support the person’s mass in an accelerating (or decelerating) frame. Mathematically, when accelerating vertically,  
$$
\text{Apparent weight} = N = m(g \pm a)
$$  
- Use $$+$$ if the acceleration is upward (increasing the reading),  
- Use $$-$$ if the acceleration is downward (decreasing the reading).

---

**34.3.2 – Apparent Weight of a 60 kg Man Accelerating Downwards at $$2 \, \text{m/s}^2$$ in an Elevator**  
Given:  
- Mass, $$m = 60 \, \text{kg}$$  
- Downward acceleration, $$a = 2 \, \text{m/s}^2$$  
- Acceleration due to gravity, $$g \approx 9.8 \, \text{m/s}^2$$

Since the elevator accelerates downward, the normal force (apparent weight) is  
$$
N = mg - ma
$$
Substitute the given values:  
$$
N = 60 \times (9.8 - 2)
$$
$$
N = 60 \times 7.8
$$
$$
N = 468 \, \text{N}
$$

---

**34.3.3 – When Does a Person Experience a Feeling of Weightlessness?**  
A person feels weightless when the normal force acting on them is zero. This occurs when the person is in free fall, i.e., when the acceleration of the system is equal to the acceleration due to gravity downward. In such a case,  
$$
N = mg - m g = 0
$$
Thus, weightlessness is experienced when the only acceleration is due to gravity and there is no supporting normal force, as in free fall.

---

**Question 35 – Two Blocks of Masses $$12 \, \text{kg}$$ and $$20 \, \text{kg}$$ on a Frictionless Surface with a Force of $$40 \, \text{N}$$ Applied to the Larger Block**

1. **Determine the common acceleration:**  
   Since the blocks are in contact and the surface is frictionless, after the force is applied, they move together. The total mass is  
   $$
   m_{\text{total}} = 12 \, \text{kg} + 20 \, \text{kg} = 32 \, \text{kg}
   $$
   The acceleration of the system is given by Newton’s second law:  
   $$
   a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{40}{32} \approx 1.25 \, \text{m/s}^2
   $$

2. **Determine the contact force between the blocks:**  
   Consider the smaller block (mass $$12 \, \text{kg}$$). The only horizontal force acting on it is the contact force $$F_c$$ from the larger block. Using Newton’s second law for this block:  
   $$
   F_c = m_{\text{small}} \times a = 12 \times 1.25 = 15 \, \text{N}
   $$
   
   Alternatively, for the larger block (mass $$20 \, \text{kg}$$), the net force acting on it is the applied force minus the contact force:
   $$
   40 - F_c = 20 \times 1.25 = 25 \, \text{N}
   $$
   Solving for $$F_c$$:
   $$
   F_c = 40 - 25 = 15 \, \text{N}
   $$
   This confirms the result.

---
**34.2.3 – Man on a Scale in an Elevator** The scale reading shows the normal force, which depends on the elevator's acceleration. If the elevator accelerates downward, the apparent weight decreases. If it accelerates upward, the apparent weight increases. --- **34.3.1 – Determining Apparent Weight** Apparent weight is determined by the normal force, which changes based on the elevator's acceleration. When accelerating downward, the apparent weight is less than the actual weight. When accelerating upward, it's more. --- **34.3.2 – Apparent Weight of a 60 kg Man Accelerating Downwards at 2 m/s²** Apparent weight = $$60 \times (9.8 - 2) = 468 \, \text{N}$$. --- **34.3.3 – When a Person Feels Weightless** A person feels weightless when the normal force is zero, which happens during free fall when the acceleration equals gravity. --- **Question 35 – Two Blocks on a Frictionless Surface** - **Total Mass:** 32 kg - **Acceleration:** 1.25 m/s² - **Contact Force Between Blocks:** 15 N

34.2.3 a man standing on a scale in
an elevator (man and scale)
34.3.1 What determines the apparent weight of a
person?

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