Problem 3 - Applied Mathematics

Question

A block of mass \(10 \mathrm{~kg}\) rests on the floor of a lift which is accelerating upwards at \(4 \mathrm{~ms}^{-2}\). The reaction of the floor of the lift on the block is: (a) \(104 \mathrm{~N}\) (b) \(96 \mathrm{~N}\) (c) \(60 \mathrm{~N}\) (d) \(30 \mathrm{~N}\) (e) \(140 \mathrm{~N}\).

Step-by-Step Solution

Answer

None of the provided options are correct; the normal force is 138 N.

Step 1: Identify Forces Acting on the Block

There are two main forces acting on the block: the gravitational force (weight) and the normal force (reaction from the floor).

Step 2: Calculate the Gravitational Force

Use the formula for the gravitational force: \[ F_{\text{gravity}} = m \times g \] where \( m = 10 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \). Therefore, \( F_{\text{gravity}} = 10 \times 9.8 = 98 \text{ N} \).

Step 3: Determine Effective Acceleration

Since the lift is accelerating upwards, the effective acceleration for the block is the sum of the gravitational acceleration and the lift's acceleration. Thus, effective acceleration \( a_{\text{eff}} = g + a = 9.8 + 4 = 13.8 \text{ m/s}^2 \).

Step 4: Calculate the Normal Force

The normal force (reaction force) can now be calculated using \[ F_{\text{normal}} = m \times a_{\text{eff}} = 10 \times 13.8 = 138 \text{ N} \].

Key Concepts

Gravitational Force

In this problem, we need to calculate the reaction of the floor of a lift on a block when the lift is accelerating upwards. First, we analyze the gravitational force, which is also referred to as weight. Weight is the force exerted by gravity on an object. For any object, we can express it mathematically using the equation: \ \( F_{\text{gravity}} = m \times g \) Here, \( m \) represents the mass of the object in kilograms (kg) and \( g \) is the acceleration due to gravity. On Earth, \( g \) typically equals 9.8 meters per second squared \( \text{m/s}^2 \).

For our specific block, which has a mass of 10 kg, the calculation is straightforward: \ \( F_{\text{gravity}} = 10 \times 9.8 = 98 \text{ N} \). Thus, the gravitational force acting on the block is 98 Newtons. Understanding this step is crucial as this force is consistent and acts downwards irrespective of the lift’s motion.

Normal Force

Next, we focus on the normal force, which is the upward reaction force from the floor of the lift on the block. This force is perpendicular to the contact surface. Typically, if an object is stationary on a horizontal surface with no additional acceleration, the normal force equals the gravitational force. However, things change when acceleration comes into play. The total normal force can be calculated with consideration to the effective acceleration (more about that later).

In this scenario where the lift is accelerating upwards, the normal force \( F_{\text{normal}} \) is actually the combined effect of the gravitational force and the force due to the lift's acceleration. We’ll calculate that by factoring in the additional upward acceleration. The correct way to represent the normal force in this context is: \ \( F_{\text{normal}} = m \times a_{\text{eff}} \).

Given that our mass \( m \) is 10 kg and the effective acceleration \( a_{\text{eff}} \) equals 13.8 \( \text{m/s}^2 \), the calculation becomes: \ \( F_{\text{normal}} = 10 \times 13.8 = 138 \text{ N} \). Hence, the upward reaction force from the floor when the lift accelerates upward is 138 Newtons.

Effective Acceleration

Effective acceleration is the crucial adjustment of acceleration in our calculation due to the lift's motion. Normally, when an object is simply at rest or moving at a constant velocity, the effective acceleration equals the gravitational acceleration \( g \) which is 9.8 \( \text{m/s}^2 \). But in our problem, since the lift accelerates upward at 4 \( \text{m/s}^2 \), this additional acceleration affects the block's perception of the gravitational pull.

We calculate effective acceleration by summing the gravitational acceleration and the lift's acceleration: \ \(a_{\text{eff}} = g + a \) where \( g \) is 9.8 \( \text{m/s}^2 \) and \( a \) is 4 \( \text{m/s}^2 \). Therefore, \ \( a_{\text{eff}} = 9.8 + 4 = 13.8 \text{ m/s}^2 \). This effective acceleration is what we use in our calculations to determine the normal force acting on the block when the lift is accelerating upward.