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Chapter 12: Problem 16

A uniform cubical crate is 0.750 m on each side and weighs 500 N. It rests on a floor with one edge against a very small, fixed obstruction. At what least height above the floor must a horizontal force of magnitude 350 N be applied to the crate to tip it?

Short Answer

Expert verified
The minimum height to apply the force is approximately 0.536 m.

Step by step solution

01

Identify the Pivot Point

The crate will tip around the edge against the obstruction, which acts as a pivot point. This is the line on which a vertical plane containing the applied force would intersect the floor.
02

Calculate the Torque by the Crate's Weight

Torque is calculated by the formula \( \tau = Fd \), where \( F \) is the force and \( d \) is the distance from the pivot point. The force due to gravity acts at the center of mass, which is at a distance of half the side of the cube from the pivot point:\[ d = \frac{0.750}{2} = 0.375 \text{ m} \]The torque due to the crate's weight then is:\[ \tau_{\text{weight}} = 500 \times 0.375 = 187.5 \text{ Nm} \].
03

Determine the Required Torque from the Horizontal Force

To tip the crate, the torque from the applied force must equal the torque due to the weight of the crate. So, if \( h \) is the height at which the force is applied:\[ \tau_{\text{force}} = 350 \times h \].
04

Solve for the Minimum Height

Set the torque produced by the horizontal force equal to the torque produced by the weight:\[ 350h = 187.5 \]Solve for \( h \):\[ h = \frac{187.5}{350} \approx 0.5357 \text{ m} \].

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Key Concepts

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

Torque Calculation
Understanding torque is essential when analyzing problems involving rotation. Torque is the measure of how much a force can cause an object to rotate about a pivot point.
Imagine torque as a twist or a rotational force. Torque depends on two main factors:
  • The magnitude of the force applied.
  • The distance from the pivot point to where the force is applied, also known as the lever arm.
If the force is applied perpendicular to the lever arm, the torque is maximal. The formula for torque is \( \tau = F \times d \), where \( \tau \) is the torque, \( F \) is the force, and \( d \) is the lever arm distance. In the crate problem, the weight creates a torque about the pivot (the obstruction's edge). The distance here is half the crate's side length. This distance plays a crucial role in determining the effectiveness of the applied force to tip the crate.
Center of Mass
Center of mass is a fundamental concept in physics and is vital to understanding balance and stability. It is the point where the total weight of an object can be considered to be concentrated.
The center of mass of symmetric and uniform bodies, like a cubical crate, is at the geometric center. Thus, the entire weight of a crate acts at this point. For our crate:
  • It is a cube, so its center of mass is halfway along each dimension.
  • The weight acts \( 0.375 \text{ m} \) from the pivot point.
Knowing the center of mass allows us to calculate how force and torque will act on the object. The balance between the weights and applied forces determines whether the crate will tip over or stay put.
Equilibrium Conditions
Understanding equilibrium helps us determine whether an object will remain motionless or move. In physics, an object is in equilibrium when the sum of all forces and the sum of all torques acting on it are both zero.
There are generally two conditions:
  • Translational Equilibrium: When the sum of all forces is zero, keeping an object in place without linear movement.
  • Rotational Equilibrium: When the sum of all torques is zero, the object doesn鈥檛 rotate.
In our crate scenario, for it to tip, the torque due to the applied force must equal the torque from the crate's weight. Solving these equilibrium equations gives us the minimum height needed for the force to successfully rotate the crate about the pivot. This balance crucially affects stability and potential motion, showcasing how forces and moments interact in real-world situations.

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Most popular questions from this chapter

After a fall, a \(95 \mathrm{~kg}\) rock climber finds himself dangling from the end of a rope that had been \(15 \mathrm{~m}\) long and \(9.6 \mathrm{~mm}\) in diameter but has stretched by \(2.8 \mathrm{~cm}\). For the rope, calculate (a) the strain, (b) the stress, and (c) the Young's modulus.

A \(75 \mathrm{~kg}\) window cleaner uses a \(10 \mathrm{~kg}\) ladder that is \(5.0 \mathrm{~m}\) long. He places one end on the ground \(2.5 \mathrm{~m}\) from a wall, rests the upper end against a cracked window, and climbs the ladder. He is \(3.0 \mathrm{~m}\) up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder does not slip. When the window is on the verge of breaking, what are (a) the magnitude of the force on the window from the ladder, (b) the magnitude of the force on the ladder from the ground, and (c) the angle (relative to the horizontal) of that force on the ladder?

A solid copper cube has an edge length of \(85.5 \mathrm{~cm}\). How much stress must be applied to the cube to reduce the edge length to \(85.0 \mathrm{~cm} ?\) The bulk modulus of copper is \(1.4 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\).

A beam of length \(L\) is carried by three men, one man at one end and the other two supporting the beam between them on a crosspiece placed so that the load of the beam is equally divided among the three men. How far from the beam's free end is the crosspiece placed? (Neglect the mass of the crosspiece.)

A trap door in a ceiling is \(0.91 \mathrm{~m}\) square, has a mass of \(11 \mathrm{~kg} .\) and is hinged along one side, with a catch at the opposite side. If the center of gravity of the door is \(10 \mathrm{~cm}\) toward the hinged side from the door's center, what are the magnitudes of the forces exerted by the door on (a) the catch and (b) the hinge?
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