Torque Due to Weight: Videos & Practice Problems

Torque is produced by an object's weight force, mg, acting at its center of gravity, which is typically the geometric center for objects with uniform mass distribution. For a fixed rod, the net torque can be calculated using the equation τ = Frsin(θ), where r is the distance from the axis of rotation and θ is the angle between the force and the lever arm. Understanding these concepts is crucial for solving physics problems involving rotational motion and equilibrium.

Torque Due to Weight

Understanding the concept of torque is essential in physics, particularly when analyzing how an object's weight can produce a torque about its axis of rotation. Torque, denoted as \( \tau \), is calculated using the formula:

\( \tau = F \cdot r \cdot \sin(\theta) \)

where \( F \) is the force applied, \( r \) is the distance from the axis of rotation to the point where the force is applied, and \( \theta \) is the angle between the force vector and the lever arm. In the context of an object like a bar or rod, the weight force \( mg \) acts at the center of gravity, which for most objects with uniform mass distribution coincides with the center of mass.

For a bar of length \( L \), the center of gravity is located at \( \frac{L}{2} \). For example, if a cylindrical rod has a mass \( m = 20 \, \text{kg} \) and a length \( L = 4 \, \text{m} \), the weight force acts at \( 2 \, \text{m} \) from the fixed end. If an additional mass \( M = 80 \, \text{kg} \) is placed at the other end of the rod, the net torque about the axis due to both weight forces must be calculated.

To find the net torque, we consider the torque produced by both masses. The torque due to the mass \( m \) is:

\( \tau_{m} = m g r_{m} \sin(\theta_{m}) \)

where \( r_{m} = 2 \, \text{m} \) (the distance to the center of mass) and \( \theta_{m} = 53^\circ \) (the angle between the force and the lever arm). For the mass \( M \) at the end of the rod, the torque is:

\( \tau_{M} = M g r_{M} \sin(\theta_{M}) \)

where \( r_{M} = 4 \, \text{m} \) and \( \theta_{M} = 53^\circ \). The gravitational acceleration \( g \) is typically approximated as \( 10 \, \text{m/s}^2 \).

Calculating these torques, we find:

\( \tau_{m} = 20 \cdot 10 \cdot 2 \cdot \sin(53^\circ) \approx 313 \, \text{N m} \)

\( \tau_{M} = 80 \cdot 10 \cdot 4 \cdot \sin(53^\circ) \approx 2504 \, \text{N m} \)

Both torques are positive as they cause counterclockwise rotation. Therefore, the net torque \( \tau_{net} \) is the sum of the individual torques:

\( \tau_{net} = \tau_{m} + \tau_{M} = 313 + 2504 = 2817 \, \text{N m} \)

This example illustrates the importance of understanding how weight forces contribute to torque in rotational dynamics. Recognizing the center of gravity and the angles involved is crucial for accurate calculations in various physics problems.

Two kids play on a seesaw that has mass 20 kg, length 3 m, and its fulcrum at its mid-point. The seesaw is originally horizontal, when the kids sit at the edge of opposite ends (m,_LEFT = 25 kg, m,_RIGHT = 30 kg). Calculate the Net Torque from the 3 weights acting on the seesaw, immediately after the kids sit (simultaneously) on their respective places.

• A. 75 N•m clockwise
• B. 75 N•m counterclockwise
• C. 150 N•m clockwise
• D. 150 N•m counterclockwise

Answer: (A)

A guy standing straight up stretches out his arm horizontally while holding a 60 lb (27.2 kg) barbell. His arm is 64 cm long and weighs 45 N. Calculate the Net Torque that the barbell and the weight of his arm produce about his shoulder. You may assume that his arm has uniform mass distribution.

• A. −156 N•m
• B. −171 N•m
• C. −185 N•m
• D. −312 N•m

Answer: (C)

Here’s what students ask on this topic:

What is torque due to weight and how is it calculated?

Torque due to weight is the rotational force exerted by an object's weight acting at its center of gravity. It is calculated using the equation:

$\tau =Fr\sin \left(\theta \right)$

where $F$ is the force (weight, $\mathrm{mg}$), $r$ is the distance from the axis of rotation to the point where the force acts, and $\theta$ is the angle between the force and the lever arm. This concept is crucial for understanding rotational motion and equilibrium in physics.

How does the center of gravity affect torque due to weight?

The center of gravity is the point where the weight of an object acts. For objects with uniform mass distribution, the center of gravity is at the geometric center. The torque due to weight is calculated by considering the force acting at this point. If the center of gravity is not at the geometric center, the distance $r$ in the torque equation changes, affecting the torque value. Understanding the center of gravity is essential for accurately calculating torque in various physics problems.

What is the significance of the angle θ in calculating torque due to weight?

The angle $\theta$ in the torque equation $\tau =Fr\sin \left(\theta \right)$ is the angle between the force vector and the lever arm. It determines the effectiveness of the force in producing rotational motion. A force applied perpendicular to the lever arm ($\theta =90°$) produces maximum torque, while a force applied parallel to the lever arm ($\theta =0°$) produces no torque. Correctly identifying this angle is crucial for accurate torque calculations.

How do you calculate the net torque on a rod with uniform mass distribution?

To calculate the net torque on a rod with uniform mass distribution, you need to consider the torques produced by all forces acting on the rod. For a rod fixed at one end, the torque due to its own weight is calculated at its center of gravity (geometric center). If additional weights are applied, calculate their torques separately and sum them up. The net torque is the sum of individual torques:

$\tau =\tau 1_{}+\tau 2_{}+...+\tau n_{}$

Ensure to consider the direction (positive or negative) of each torque based on the rotational direction they cause.

What is the difference between center of gravity and center of mass?

The center of gravity is the point where the gravitational force acts on an object, while the center of mass is the point where the mass of an object is concentrated. For most practical purposes in a uniform gravitational field, these two points coincide. However, in a non-uniform gravitational field, they can differ. In physics problems involving torque due to weight, the center of gravity is typically used, especially when dealing with uniform mass distribution.