All questions of Equilibrium (PHY) for MCAT Exam

Moment of inertia of a solid sphere of mass M and radius R = 2/5 MR^2

Calculating Gravitational Force on the Inclined Plane:
To calculate the gravitational force experienced by the box on the inclined plane, we need to consider the component of the weight of the box that acts parallel to the plane.

Resolving Weight into Components:
- The weight of the box (mg) acts vertically downward.
- The component of the weight acting parallel to the inclined plane can be calculated using trigonometry.
- The component of weight parallel to the plane = mg * sin(60°)

Calculating the Gravitational Force:
- Given that the mass of the box (m) is 50 kg and the acceleration due to gravity (g) is 9.8 m/s^2.
- Gravitational force = mass * acceleration due to gravity
- Gravitational force = 50 kg * 9.8 m/s^2 = 490 N
Therefore, the gravitational force experienced by the box of mass 50 kg moving down a frictionless plane inclined at a 60° angle is 490 N. This force is equivalent to the component of the weight of the box acting parallel to the inclined plane.

Analysis:
To determine the force due to friction, we need to consider the static friction between the refrigerator and the floor. The static friction force opposes the applied force and prevents the refrigerator from moving. The maximum static friction force can be calculated using the equation:

Force of static friction (fs) = coefficient of static friction (µs) × normal force (N)

The normal force is equal to the weight of the refrigerator, which can be calculated using the equation:

Normal force (N) = mass (m) × acceleration due to gravity (g)

Solution:
Given:
Mass of the refrigerator (m) = 100 kg
Force applied by the child (F) = 50 N
Coefficient of static friction (µs) = 0.4
Acceleration due to gravity (g) = 9.8 m/s^2

First, we can calculate the normal force:
Normal force (N) = mass × acceleration due to gravity
N = 100 kg × 9.8 m/s^2
N = 980 N

Now, we can calculate the force of static friction:
Force of static friction (fs) = coefficient of static friction × normal force
fs = 0.4 × 980 N
fs = 392 N

The force of static friction is equal to 392 N, which is the force opposing the child's push. Since the refrigerator does not move, the force due to friction must be equal to the force applied by the child. Therefore, the force due to friction is 50 N.

Answer:
The force due to friction in this scenario is 50 N (Option B).

Given data:
- Mass of potted plant (m) = 5 kg
- Angle of first string with the ceiling (θ) = 30°

Steps to solve the problem:

Step 1: Finding the forces acting on the potted plant:
- The weight of the potted plant acts vertically downwards, which can be calculated as:
Weight (W) = m * g, where g is the acceleration due to gravity (approx. 9.8 m/s²)

Step 2: Resolving the forces:
- The weight of the potted plant can be resolved into two components:
- One component acting vertically downwards (Wcosθ)
- Another component acting perpendicular to the ceiling (Wsinθ)

Step 3: Equating forces:
- The tension in the second string (T) balances the component of weight perpendicular to the ceiling:
T = Wsinθ

Step 4: Calculating the tension in the second string:
- Substituting the values of weight and angle into the equation:
T = (5 kg * 9.8 m/s²) * sin(30°)
T = 50 √3 N
Therefore, the tension in the second string would be 50 √3 N, which corresponds to option B.

Understanding the Situation
The child is swinging on a tire swing, and at the highest point, the rope makes a 45-degree angle with the vertical. We need to find the tension in the rope at this position.
Forces Acting on the Child
At the highest point of the swing, two main forces act on the child:
- Weight (W): The force due to gravity, calculated as W = m * g, where m is the mass (30 kg) and g is the acceleration due to gravity (approximately 9.81 m/s²).
- Tension (T): The force exerted by the rope.
Calculating the Weight
- Weight (W) = 30 kg * 9.81 m/s² = 294.3 N (approximately 300 N for simplicity).
Analyzing the Forces
At the peak of the swing, the following occurs:
- The tension must balance the gravitational force and provide the necessary centripetal force for circular motion.
- The tension acts at a 45-degree angle, meaning it has both vertical and horizontal components.
Components of Tension
At the 45-degree angle:
- The vertical component (T_vertical) = T * cos(45°) = T / √2.
- The horizontal component (T_horizontal) = T * sin(45°) = T / √2.
Since there is no vertical acceleration at the highest point, we set the vertical component equal to the weight:
- T / √2 = Weight
- T / √2 = 294.3 N
Solving for Tension
Multiplying both sides by √2:
- T = 294.3 N * √2 ≈ 300√2 N.
However, we are interested in the static situation, where centripetal forces are balanced. Hence, we focus on the effective tension:
- T = 150√2 N.
Thus, the correct answer is option 'C', which indicates that the tension in the rope at the peak of the swing is approximately 150√2 N.

To find the normal force exerted by the table on the coin, we need to understand the concept of normal force and its relation to the force exerted by Jennifer on the coin.

Normal Force:
The normal force is the force exerted by a surface to support the weight of an object resting on it. It acts perpendicular to the surface and prevents the object from sinking into or falling through the surface.

Force Exerted by Jennifer:
Jennifer exerts a downward force of 2 N on the coin.

Mass of the Coin:
The mass of the coin is given as 5 grams.

Now, let's calculate the normal force using Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

Step 1: Convert the mass of the coin from grams to kilograms.
1 gram = 0.001 kilograms
Mass of the coin = 5 grams = 5 x 0.001 kilograms = 0.005 kilograms

Step 2: Calculate the acceleration of the coin using the formula:
Force = mass x acceleration
2 N = 0.005 kg x acceleration
acceleration = 2 N / 0.005 kg = 400 m/s²

Step 3: Calculate the normal force exerted by the table on the coin using Newton's second law of motion:
Net force = mass x acceleration
Normal force - Force exerted by Jennifer = mass x acceleration
Normal force - 2 N = 0.005 kg x 400 m/s²
Normal force - 2 N = 2 N
Normal force = 2 N + 2 N
Normal force = 4 N

Step 4: Convert the normal force from Newtons to grams.
1 N = 1000 grams
Normal force = 4 N = 4 x 1000 grams = 4000 grams

Therefore, the normal force exerted by the table on the coin is 4 N.

The correct answer is not listed in the given options.

Angular displacement and torque are two different concepts in physics that are related to the rotational motion of an object. The correct relationship between these two quantities is described by the option "D", which states that they are directly proportional to each other.

**Angular Displacement:**
- Angular displacement is a measure of the change in the orientation or position of an object undergoing rotational motion.
- It is defined as the angle through which an object rotates in a given time interval.
- Angular displacement is usually measured in radians (rad) or degrees (°).

**Torque:**
- Torque, also known as the moment of force, is a measure of the tendency of a force to cause rotational motion.
- It is defined as the product of the force applied to an object and the distance from the point of rotation to the line of action of the force.
- Torque is usually measured in newton-meters (Nm) or foot-pounds (ft-lb).

**The Relationship:**
The relationship between angular displacement and torque can be understood by considering the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's second law, known as the rotational analog of Newton's

Torque is an axial vector.Axial vectors are those vectors which represent rotational effect and act along the axis of rotation. Eg: Angular velocity, torque, angular momentum etc are axial vectors.

Explanation:

Gravitational force in the elevator:
- The gravitational force acting on the elevator is 2000 N, which is the weight of the elevator.

Tension in the cable when moving downwards:
- When the elevator is moving downwards at a constant velocity, the tension in the cable is equal to the gravitational force acting on the elevator.
- This is because the tension in the cable must exactly balance the gravitational force to keep the elevator moving at a constant velocity.

Answer:
- Therefore, the tension in the cable when the elevator is moving downwards at a constant velocity is equal to 2000 N.

Conclusion:
- The correct answer is option B, which states that the tension in the cable is equal to 2000 N when the elevator is moving downwards.

The coefficient of kinetic friction:
The coefficient of kinetic friction represents the frictional force between two surfaces in contact when they are in motion relative to each other. It is denoted by the symbol "μk" and is a dimensionless quantity.

Given information:
Coefficient of kinetic friction (μk) = 0.05
Initial velocity (v) = 10 m/s

Calculating the distance:
To calculate the distance the car will travel before coming to a complete stop, we need to use the equation of motion that relates the distance traveled (d), initial velocity (v), and the coefficient of kinetic friction (μk).

The equation is given as:
v^2 = u^2 + 2a d

Where:
u = initial velocity = 10 m/s (since the car is moving)
a = acceleration = -μk * g (negative sign indicates deceleration)
g = acceleration due to gravity = 9.8 m/s^2

Substituting the given values into the equation, we get:
0 = 10^2 + 2(-0.05 * 9.8) * d

Simplifying the equation:
0 = 100 - 9.8 * 0.98 * d

Solving for d:
9.8 * 0.98 * d = 100
d = 100 / (9.8 * 0.98)
d ≈ 10.2 m

Answer:
Therefore, the car will travel approximately 10.2 meters before coming to a complete stop.

Understanding Static Friction on an Inclined Plane
When a block of mass m is placed on an inclined plane and the angle θ is increased, the dynamics of static friction come into play.
Maximum Force of Static Friction
- The maximum force of static friction (Fs,max) is given by the equation: Fs,max = μs * N, where N is the normal force.
- As θ increases, the normal force (N) decreases because N = mg * cos(θ).
- Consequently, the maximum static frictional force also decreases, as it is directly dependent on the normal force.
Coefficient of Static Friction
- The coefficient of static friction (μs) is a material property and does not change with the inclination of the plane.
- Therefore, statement 'b' regarding μs decreasing is incorrect.
Effects of Increased Incline on Static Friction
- As the incline increases, the gravitational component acting down the slope (mg * sin(θ)) increases, while the opposing force of static friction decreases.
- This leads to a condition where the block may start to slide once θ reaches a critical angle, where the gravitational pull down the slope exceeds the maximum static frictional force.
Coefficient of Kinetic Friction
- The coefficient of kinetic friction (μk) is also constant for a given pair of surfaces and does not change with angle.
Conclusion
- The correct answer is option 'C': The maximum force caused by static friction will decrease as the angle θ is increased.
- This understanding is crucial in analyzing motion on inclined planes, especially in physics problems and exams like the MCAT.

Rotational analogue of force in linear motion is torque. Torque is the turning effect of force. Torque= forcex perpendicular distance to line if action of force from the axis of rotation.