When only the front wheels of an automobile are on a platform scale, the scale balances at \(8.0 \mathrm{kN} ;\) when only the rear wheels are on the scale, it balances at \(6.0 \mathrm{kN}\). What is the weight of the automobile, and how far is its center of mass behind the front axle? The distance between the axles is \(2.8 \mathrm{~m}\).

Short Answer

Expert verified
Answer: The weight of the automobile is 14 kN and its center of mass is 1.12 m behind the front axle.

Step by step solution

01

Define the variables and set up force balance equation

Let's define the following variables: - \(W_F\): force on the front wheels (given as \(8.0\,\text{kN}\)) - \(W_R\): force on the rear wheels (given as \(6.0\,\text{kN}\)) - \(W_{A}\): total weight of the automobile - \(x\): distance from the center of mass to the front axle - \(L\): distance between the axles (given as \(2.8\,\text{m}\)) To set up the force balance equation, we need to account for the forces acting on the car: $$W_{A} = W_F + W_R$$ Plugging in the given values for \(W_F\) and \(W_R\), we get: $$W_{A} = 8.0\,\text{kN} + 6.0\,\text{kN}$$ Now, we can calculate the weight of the automobile: $$W_{A} = 14\,\text{kN}$$
02

Set up the moment balance equation

We will set up the moment balance equation about the front axle. The moments due to the forces from the front wheels, rear wheels, and weight of the automobile must all balance. The equation will look like this: $$W_{A}x = W_FL + W_R(L - x)$$
03

Substitute the known values and solve for \(x\)

Now, let's substitute the known values into the moment balance equation: $$14\,\text{kN}x = 8.0\,\text{kN}(2.8\,\text{m}) + 6.0\,\text{kN}(2.8\,\text{m} - x)$$ Simplify the equation and isolate \(x\): $$14x = 22.4 - 6x$$ $$20x = 22.4$$ $$x = 1.12\,\text{m}$$
04

State the final answers

The weight of the automobile is \(14\,\text{kN}\) and its center of mass is \(1.12\,\text{m}\) behind the front axle.

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Key Concepts

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

Moment Balance Equation
Understanding the moment balance equation is crucial for problems involving rotational equilibrium and finding the center of mass in physics. This concept is based on the principle that the sum of all moments about any point is zero when the system is in equilibrium. In the case of our automobile problem, we consider the moments produced by the forces acting on the front and rear wheels.

The moment balance equation is mathematically expressed as \( W_Ax = W_FL + W_R(L - x) \). Hier, \(W_A\) represents the total weight of the automobile, \( x \) is the distance from the center of mass to the front axle, \( L \) is the axle separation distance, \( W_F \) and \( W_R \) are the weights supported by the front and rear wheels, respectively. By selecting the front axle as the pivot point, we eliminate the moment due to the force on the front wheels, simplifying our calculations. Balancing moments allow us to solve for the unknown variable \( x \) which reflects the location of the center of mass relative to the front axle.
Force Balance
Force balance is another fundamental concept in physics that deals with the equilibrium of forces acting on a body. In a state of balance, the sum of all forces acting on an object is zero. Applying this to our automobile scenario, the total force balance equation is represented as \( W_A = W_F + W_R \).

This equation means that the total weight of the car, \( W_A \), is the sum of the weights on the front and rear wheels. If we simply add up the known forces of \( 8.0 \) kN and \( 6.0 \) kN from the problem statement, we can straightforwardly deduce that the car's total weight is \( 14 \) kN. Recognizing the force balance helps us not just solve for the weight, but it also reinforces our understanding of how forces interact and combine in physical systems.
Physics Problem Solving
Physics problem solving often involves a systematic approach to dissect complicated situations into manageable parts. By defining variables, setting up relevant equations, and employing appropriate principles, we can navigate through complex problems. This procedure is exemplified in the steps taken to solve for the automobile’s weight and the center of mass position.

In our example, we've meticulously defined variables like forces on the wheels (\(\ W_F \) and \(\ W_R \)), total weight (\(\ W_A \)), distance from the center of mass to the front axle (\(\ x \)), and the distance between axles (\(\ L \)). By setting up force balance and moment balance equations and consistently substituting known values, the unknowns are progressively uncovered. Following these steps carefully allows students to build a solid foundation for analytical thinking in physics and develop a reliable strategy for approaching a wide range of physics problems.

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Most popular questions from this chapter

A \(3.0-\mathrm{kg}\) broom is leaning against a coffee table. \(A\) woman lifts the broom handle with her arm fully stretched so that her hand is a distance of \(0.45 \mathrm{~m}\) from her shoulder What torque is produced on her shoulder if her arm is at an angle of \(50^{\circ}\) below the horizontal? a) \(7.0 \mathrm{~N} \mathrm{~m}\) c) \(8.5 \mathrm{~N} \mathrm{~m}\) b) \(5.8 \mathrm{~N} \mathrm{~m}\) d) \(10.1 \mathrm{~N} \mathrm{~m}\)

The angular displacement of a torsional spring is proportional to the applied torque; that is \(\tau=\kappa \theta,\) where \(\kappa\) is a constant. Suppose that such a spring is mounted to an arm that moves in a vertical plane. The mass of the arm is \(45.0 \mathrm{~g}\), and it is \(12.0 \mathrm{~cm}\) long. The arm-spring system is at equilibrium with the arm at an angular displacement of \(17.0^{\circ}\) with respect to the horizontal. If a mass of \(0.420 \mathrm{~kg}\) is hung from the arm \(9.00 \mathrm{~cm}\) from the axle, what will be the angular displacement in the new equilibrium position (relative to that with the unloaded spring)?

A uniform, equilateral triangle of side length \(2.00 \mathrm{~m}\) and weight \(4.00 \cdot 10^{3} \mathrm{~N}\) is placed across a gap. One point is on the north end of the gap, and the opposite side is on the south end. Find the force on each side.

An SUV has a height \(h\) and a wheelbase of length b. Its center of mass is midway between the wheels and at a distance \(\alpha h\) above the ground, where \(0<\alpha<1\). The SUV enters a turn at a dangerously high speed, \(v\). The radius of the turn is \(R(R \gg b)\), and the road is flat. The coefficient of static friction between the road and the properly inflated tires is \(\mu_{s}\). After entering the turn, the SUV will either skid out of the turn or begin to tip. a) The SUV will skid out of the turn if the friction force reaches its maximum value, \(F \rightarrow \mu_{\mathrm{s}} N\). Determine the speed, \(v_{\text {skid }},\) for which this will occur. Assume no tipping occurs. b) The torque keeping the SUV from tipping acts on the outside wheel. The highest value this force can have is equal to the entire normal force. Determine the speed, \(v_{\text {tip }}\), at which this will occur. Assume no skidding occurs. c) It is safer if the SUV skids out before it tips. This will occur as long as \(v_{\text {skid }}

In the absence of any symmetry or other constraints on the forces involved, how many unknown force components can be determined in a situation of static equilibrium in each of the following cases? a) All forces and objects lie in a plane. b) Forces and objects are in three dimensions. c) Forces act in \(n\) spatial dimensions.

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