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Chapter 11: Problem 17

You take your dog Clea to the vet, and the doctor decides he must locate the little beast's center of gravity. It would be awkward to hang the pooch from the ceiling, so the vet must devise another method. He places Clea's front feet on one scale and ber hind feet on another. The front scale reads \(157 \mathrm{N},\) while the rear scale reads 89 \(\mathrm{N}\) . The vet next measures Clea and finds that her rear feet are 0.95 \(\mathrm{m}\) behind her front feet. How much does Clea weigh, and where is her center of gravity?

Short Answer

Expert verified
Clea weighs 246 N; her center of gravity is 0.538 m from her front feet.

Step by step solution

01

Understand Total Weight Calculation

The total weight of Clea can be calculated by summing the readings from both scales since the scales support the entire weight of Clea. \[ W = W_{front} + W_{rear} \]Where \(W_{front} = 157 \mathrm{N} \), the front scale reading, and \(W_{rear} = 89 \mathrm{N} \), the rear scale reading.
02

Calculate Total Weight

Add the front and rear scale readings to get Clea's total weight. \[ W = 157 \mathrm{N} + 89 \mathrm{N} = 246 \mathrm{N} \]Thus, Clea weighs 246 Newtons.
03

Analyze Center of Gravity

The center of gravity can be found by considering the balance of moments. We set the moment about the front feet (where the front scale is) equal to the moment about the center of gravity:\[ M_{rear} = W_{rear} \times d = (W_{total} - W_{rear}) \times x \]where- \( d = 0.95 \mathrm{m} \), distance between front and rear feet,- \( x \), distance from the front feet to the center of gravity.
04

Rewrite the Equation for Center of Gravity

Substitute the known values into the equation:\[ 89 \mathrm{N} \times 0.95 \mathrm{m} = (246 \mathrm{N} - 89 \mathrm{N}) \times x \]Simplify this to:\[ 84.55 = 157 \times x \]
05

Solve for Center of Gravity

Now solve the equation for \( x \) to find the distance from the front feet to Clea's center of gravity:\[ 84.55 = 157x \]\[ x = \frac{84.55}{157} \]\[ x \approx 0.538 \mathrm{m} \]
06

Conclusion

Clea's center of gravity is approximately 0.538 meters behind her front feet, and her total weight is 246 Newtons.

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Key Concepts

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

Weight Calculation
Understanding how to calculate weight is essential in many physics problems. In this problem, calculating Clea's total weight is relatively straightforward. Clea's weight is distributed between two scales, one under her front feet and one under her rear. These scales measure the force she's exerting downwards due to gravity. To find her total weight, you simply sum up the forces recorded by the scales. Clea's front feet have a reading of 157 Newtons and her rear feet show 89 Newtons. Therefore, by adding these two values, you determine that Clea weighs:
  • 157 N + 89 N = 246 N
This method is simple yet effective for finding total weight when multiple support points are involved in the measurement.
Moments Balance
To find out where Clea's center of gravity (CG) lies, we employ the principle of moments balance. This is a crucial concept in physics when dealing with systems that are not in motion. Moments (or torques) are the turning forces around a pivot point, and they need to be balanced for an object to be in equilibrium.In Clea's case, the equilibrium condition involves the forces acting around her center of gravity. The total moment around her front feet must equal the total moment around her CG. This is expressed in the equation:\[ M_{rear} = W_{rear} \times d = (W_{total} - W_{rear}) \times x \]Here, the value of \( d \) (the distance between the front and rear feet) is 0.95 meters. By balancing these moments, we can find how far the center of gravity is from her front feet.
Physics Problem Solving
Physics problem solving often involves breaking down complex real-world situations into simpler, manageable parts. In solving for Clea's weight and center of gravity, you start by understanding the task and breaking it down into systematic steps.
  • Identify what needs to be calculated: Clea's total weight and center of gravity.
  • Gather known values: front and rear scale readings, plus the distance between her feet.
  • Write down the relevant formulas: one for calculating total weight and another for moments balance.
  • Substitute values into the equations and solve step-by-step: this process ensures you get accurate results.
Successful physics problem solving relies on logical reasoning and a clear understanding of the underlying principles, like weights and moments in this case.
Newton's Laws of Motion
Isaac Newton's Laws of Motion are foundational principles in physics, providing a framework for understanding how forces influence motion and equilibrium. In our problem with Clea, an implicit understanding of Newton's first law is essential. This law states that an object remains at rest or in uniform motion unless acted upon by a force. In this situation, Clea is at rest, which means the forces (her weight distributed over the two scales) are in equilibrium. The scales give us the force readings because Clea is not accelerating. Thus, her weight ( 246 N otal force) results from gravity acting upon her mass, demonstrating the balance described by Newton's laws. By applying these principles, we not only determine her weight but also gain insight into where her center of gravity is, as the distribution of force (weight) influences this location within the confines of her body.

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Most popular questions from this chapter

Bulk Modulus of an Ideal Gas. The equation of state (the equation relating pressure, volume, and temperature) for an ideal gas is \(p V=n R T\) , where \(n\) and \(R\) are constants. (a) Show that if the gas is compressed while the temperature \(T\) is held constant, the bulk modulus is equal to the pressure. (b) When an ideal gas is compressed without the transfer of any heat into or out of it, the pressure and volume are related by \(p V^{\gamma}=\) constant, where \(\gamma\) is a constant having different values for different gases. Show that, in this case, the bulk modulus is given by \(B=\gamma p\) .

In a materials testing laboratory, a metal wire made from a new alloy is found to break when a tensile force of 90.8 \(\mathrm{N}\) is applied perpendicular to each end. If the diameter of the wire is \(1.84 \mathrm{mm},\) what is the breaking stress of the alloy?

A \(15,000-N\) crane pivots around a friction-free axle at its base and is supported by a cable making a \(25^{\circ}\) angle with the crane (Fig. 11.29 ). The crane is 16 \(\mathrm{m}\) long and is not uniform, its center of gravity being 7.0 \(\mathrm{m}\) from the axle as measured along the crane. The cable is attached 3.0 \(\mathrm{m}\) from the upper end of the crane. When the crane is raised to \(55^{\circ}\) above the horizontal holding an \(11,000-N\) pallet of bricks by a 2.2 -in very light cord, find (a) the tension in the cable, and \((b)\) the horizontal and vertical components of the force that the axle exerts on the crane. Start with a free-body diagram of the crane.

Shear forces are applied to a rectangular solid. The same forces are applied to another rectangular solid of the same material, but with three times each edge length. In each case the forces are small enough that Hooke's law is obeyed. What is the ratio of the shear strain for the larger object to that of the smaller object?

Two people are carrying a uniform wooden board that is 3.00 \(\mathrm{m}\) long and weighs 160 \(\mathrm{N}\) . If one person applies an upward force equal to 60 \(\mathrm{N}\) at one end, at what point does the other person lift? Begin with a free-body diagram of the board.
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