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Chapter 12: Problem 16

A uniform cubical crate is \(0.750 \mathrm{~m}\) on each side and weighs \(500 \mathrm{~N}\), 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 \(350 \mathrm{~N}\) be applied to the crate to tip it?

Short Answer

Expert verified
The force must be applied at a height of 0.536 m.

Step by step solution

01

Understanding the Problem

We have a cubical crate resting on the floor, which is being tipped over a fixed obstruction by applying a horizontal force. The crate has a weight of \(500 \, \text{N}\) and dimensions of \(0.750 \, \text{m} \times 0.750 \, \text{m} \times 0.750 \, \text{m}\). The obstruction causes the crate to pivot around the edge at the base. We need to find the minimum height above the floor where a \(350 \, \text{N}\) horizontal force can be applied to just start tipping the crate.
02

Analyzing the Moment of Force

The tipping will occur when the moment created by the applied force about the pivot point equals the resisting moment due to the crate's weight. The moment arm for the weight will be half the diagonal of the cube's bottom face, \(0.750/2\), because the cube rotates about the lower edge. The height \(h\) at which the force is applied is what we need to find.
03

Calculating the Moment Due to the Crate's Weight

The resisting moment due to the crate's weight \(W = 500 \, \text{N}\) acts at the center of the crate, which is at a distance of \(0.375 \, \text{m}\) horizontally from the pivot (half of one side). Thus, the moment due to weight is \(500 \, \text{N} \times 0.375 \, \text{m}\).
04

Calculating the Moment Due to the Applied Force

The required tipping moment due to the applied force \(F = 350 \, \text{N}\) should equal the resisting moment. If it acts at a height \(h\), then the moment is \(350 \, \text{N} \times h\). We set this equal to the moment due to the weight from Step 3.
05

Solving for the Required Height

Set the moments equal to each other: \(350 \, \text{N} \times h = 500 \, \text{N} \times 0.375 \, \text{m}\).Solving for \(h\):\[ h = \frac{500 \times 0.375}{350} = \frac{187.5}{350} = 0.5357 \, \text{m} \]
06

Conclusion

The minimum height above the floor at which the \(350 \, \text{N}\) force must be applied to tip the crate is approximately \(0.536 \, \text{m}\).

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Key Concepts

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

Moments and Torque
To fully grasp how moments and torque work in this scenario, let's delve into the concepts. A moment, often referred to as torque, is essentially a measure of the "rotational force" about a specific point. In this scenario, it's about how much "twisting" effect is applied to cause the crate to tip over from its resting position.
  • The "pivot point" is crucial; it is the fixed obstruction at the crate's base in this context.
  • Torque is calculated by multiplying the force applied by the distance (moment arm) from the pivot point to where the force is applied.
  • When the applied force's moment equals the resisting moment (due to the crate's weight), this balance allows for tipping.
Understanding these points offers a basis to predict how forces will affect rotational motion around a pivot.
Equilibrium Conditions
For the crate to initially stay put, it must be in a state of equilibrium. There are two types of equilibrium to consider here: static and dynamic. Static equilibrium is crucial in this example.
  • Static equilibrium occurs when the sum of all forces and the sum of all torques (or moments) acting on an object are zero.
  • The crate, at rest, is in static equilibrium until the applied force creates a moment large enough to overcome the resisting moment, causing tipping.
  • Analyzing equilibrium helps in determining when the applied force is just sufficient to tip the crate without additional sliding or shifting.
These conditions underline the application of forces and predict the subsequent motion if these criteria are altered.
Tipping Force Calculation
To calculate the required force or its point of application to cause the crate to tip, we use the concept of moments. Here, the force of 350 N is applied horizontally at a height that needs determining.
  • We know the crate's weight creates a resisting moment around the pivot, calculated using the weight's distance from the edge (0.375 m in this case).
  • The applied moment needs to match this resistance for tipping to start.
  • By setting up the equation with these known values, \( 350 \, \text{N} \times h = 500 \, \text{N} \times 0.375 \, \text{m} \), we solve for \( h \), the minimum height for applying the force.
  • Calculating this gives \( h \approx 0.536 \, \text{m} \), indicating the precise height the force must be applied to achieve equilibrium tipping conditions.
These calculations showcase how physics principles and formulas such as torque are applied to solve real-world problems.

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Most popular questions from this chapter

A uniform ladder is \(10 \mathrm{~m}\) long and weighs \(200 \mathrm{~N}\). In Fig. \(12-76\), the ladder leans against a vertical, frictionless wall at height \(h=8.0 \mathrm{~m}\) above the ground. A horizontal force \(\vec{F}\) is applied to the ladder at distance \(d=2.0 \mathrm{~m}\) from its base (measured along the ladder). (a) If force magnitude \(F=50 \mathrm{~N}\), what is the force of the ground on the ladder, in unit-vector notation? (b) If \(F=150 \mathrm{~N}\), what is the force of the ground on the ladder, also in unit-vector notation? (c) Suppose the coefficient of static friction between the ladder and the ground is \(0.38 ;\) for what minimum value of the force magnitude \(F\) will the base of the ladder just barely start to move toward the wall?

A rope of negligible mass is stretched. horizontally between two supports that are \(3.44\) \(\mathrm{m}\) apart. When an object of weight \(3160 \mathrm{~N}\) is hung at the center of the rope, the rope is observed to sag by \(35.0 \mathrm{~cm} .\) What is the tension in the rope?

An archer's bow is drawn at its midpoint until the tension in the string is equal to the force exerted by the archer. What is the angle between the two halves of the string?

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 \(1.4 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2} .\)

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.
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