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Chapter 7: Problem 60

A uniform cube of side \(a\) and mass \(m\) rests on a rough horizontal table. A horizontal force \(F\) is applied normal to one of the faces at a point that is directly above the centre of face, at a height \(\frac{3 a}{4}\) above the base. The minimum value of \(F\) for which the cube begins to tilt about the edge is (Assume that the cube does not slide) (a) \(\frac{m g}{4}\) (b) \(\frac{2 m g}{3}\) (c) \(\frac{3 m g}{4}\) (d) \(m g\)

Short Answer

Expert verified
The minimum force is \( \frac{2mg}{3} \), option (b).

Step by step solution

01

Determine the Torque due to Force F

The force \( F \) is applied at a height \( \frac{3a}{4} \) above the base of the cube. Torque \( \tau \) due to force \( F \) about the edge of the cube (where it starts to tilt) is given by: \[ \tau_F = F \cdot \frac{3a}{4} \] This expression determines the turning effect of the force \( F \) about the edge.
02

Determine the Torque due to Gravity

The weight \( mg \) of the cube acts through its center, which is at a distance \( \frac{a}{2} \) from the tilting edge horizontally because the cube is symmetric. The torque \( \tau_g \) about the edge due to gravity is: \[ \tau_g = m g \cdot \frac{a}{2} \] This torque opposes the torque due to \( F \).
03

Equate Torques to Find Minimum F

For the cube to just begin to tilt, the torque due to force \( F \) should balance the torque due to gravity. Hence: \[ F \cdot \frac{3a}{4} = m g \cdot \frac{a}{2} \] Simplifying this equation: \[ F = m g \cdot \frac{a}{2} \cdot \frac{4}{3a} = \frac{2mg}{3} \] Thus, the minimum force \( F \) for which the cube begins to tilt is obtained.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Cube Mechanics
Imagine a cube sitting on a table. It has an equal length on all sides, which we denote as side length \( a \). Because the cube is uniform, its mass is evenly spread throughout its volume, which has implications on how it behaves under force.
The center of mass of the cube is right in the middle due to its uniform nature. This is where we assume all the weight \( mg \) acts from, making calculations easier and more intuitive. This uniformity allows us to use simplified physics equations to predict how it will move or tumble when forces are applied.
Understanding these mechanics is crucial for determining how forces such as gravity and applied forces influence the cube鈥檚 motion and equilibrium. The symmetry makes calculations predictable and straightforward.
Horizontal Force Application
When a force is applied in the real world, it often acts in a particular direction and at a specific point. In this exercise, we apply a horizontal force \( F \) to the cube. This means the force pushes sideways rather than up or down, at a height \( \frac{3a}{4} \) up from the base.
This configuration of the force has significance because where you apply the force affects whether the object might slide or tilt. Here, the force is placed directly above the center of a face, impacting the balance and tendency of the cube to start tilting. Knowing how to precisely calculate this effect depends on understanding the relationship between force direction, point of application, and how they generate torque.
Tilting Threshold Calculation
To determine when a cube starts to tilt, we must compare the torques generated by different forces acting at various points. Torque is a measure of how much a force causes an object to rotate. For the cube to tilt, the torque due to the applied horizontal force \( F \) must equal the torque due to gravity acting on the cube.
Imagine the cube trying to rotate around one of its edges. The force \( F \), applied at a height \( \frac{3a}{4} \), creates one torque, while gravity, pulling down through the cube's center of mass, creates the opposing torque slightly offset from this edge.
The equation \( F \cdot \frac{3a}{4} = mg \cdot \frac{a}{2} \) shows this balance. Rearranging gives \( F = \frac{2mg}{3} \). This result tells us the minimum force needed to overcome gravity's stabilizing effect, causing the cube to start tilting rather than sliding.

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Most popular questions from this chapter

A uniform metre stick of mass \(M\) is hinged at one end and supported in a horizontal direction by a string attached to the other end. What should be the initial acceleration (in \(\left.\mathrm{rad} / \mathrm{s}^{2}\right)\) of the stick if the string is cut? (a) \(\frac{3}{2} g\) (b) \(g\) (c) \(3 g\) (d) \(4 g\)

A wheel with an initial angular velocity \(\omega_{o}\) reaches an angular velocity of \(5 \omega_{o}\) while it turns through an angle of 6 rad. Its uniform angular acceleration \(\alpha\) is (a) \(\frac{1}{3} \omega_{o}^{2} \mathrm{rad} / \mathrm{sec}^{2}\) (b) \(\frac{2}{3} \omega_{a}^{2} \mathrm{rad} / \mathrm{sec}^{2}\) (c) \(2 \omega_{o}^{2} \mathrm{rad} / \mathrm{sec}^{2}\) (d) \(4 \omega_{o}^{2} \mathrm{rad} / \mathrm{sec}^{2}\)

A circular platform is mounted on a vertical frictionless axle. Its radius is \(r=2 m\) and its moment of inertia is \(I=\) \(200 \mathrm{~kg}-\mathrm{m}^{2}\). It is initially at rest. A \(70 \mathrm{~kg}\) man stands on the edge of the platform and begins to walk along the edge at speed \(v_{0}=10 \mathrm{~m} / \mathrm{s}\) relative to the ground. The angular velocity of the platform is (a) \(1.2 \mathrm{rad} / \mathrm{s}\) (b) \(0.4 \mathrm{rad} / \mathrm{s}\) (c) \(2.0 \mathrm{rad} / \mathrm{s}\) (d) \(0.7 \mathrm{rad} / \mathrm{s}\)

A uniform thin bar of mass \(6 \mathrm{~m}\) and length \(12 L\) is bent to make a regular hexagon. Its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the hexagon is (a) \(20 \mathrm{~mL}^{2}\) (b) \(6 m L^{2}\) (c) \(\frac{12}{5} m L^{2}\) (d) \(30 \mathrm{~mL}^{2}\)

An inclined plane makes an angle of \(30^{\circ}\) with the horizontal. A solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration equal to (a) \(\frac{g}{3}\) (b) \(\frac{5 g}{7}\) (c) \(\frac{2 g}{3}\) (d) \(\frac{5 g}{14}\)
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