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Chapter 11: Problem 44

A pickup truck has a wheelbase of 3.00 \(\mathrm{m}\) . Ordinarily. \(10,780 \mathrm{N}\) rests on the front wheels and 8820 \(\mathrm{N}\) on the rear wheels when the truck is parked on a level road. (a) \(\mathrm{A}\) box weighing 3600 \(\mathrm{N}\) is now placed on the tailgate, 1.00 \(\mathrm{m}\) behind the rear axie. How much total weight now rests on the front wheels? On the rear wheels? (b) How much weight would need to be placed on the tail- gate to make the front wheels come off the ground?

Short Answer

Expert verified
(a) Front: 4,793.33 N; Rear: 18,406.67 N. (b) 15,680 N.

Step by step solution

01

Analyze the System Without the Box

Initially, the total force acting on the truck is the weight on both sets of wheels, which is given by the sum of forces on front (\(F_f\)) and rear (\(F_r\)) wheels: \( F_f + F_r = 10,780 \, \text{N} + 8,820 \, \text{N} = 19,600 \, \text{N} \). The center of gravity of the truck's weight is determined by the load distribution between front and rear wheels. The truck's wheelbase is \(3.00 \, \text{m}\).
02

Place the Box and Calculate New Torques

When a box weighing 3,600 N is placed 1.00 m behind the rear axle, it creates an additional torque on the system. Calculate the torque generated by the box with respect to the rear axle: \( \tau_{box} = 3,600 \, \text{N} \times 1.00 \, \text{m} = 3,600 \, \text{Nm} \).
03

Determine New Load on Wheels

To find the new distribution of weight, the sum of torques about the rear axle must be zero for equilibrium. The original weight distribution must adjust for the added torque to keep the system stationary. Let \(F'_f\) and \(F'_r\) be the new forces on the front and rear wheels, respectively. The balance of torque about the rear axle is: \[ F'_f \times 3.00 \, \text{m} = 10,780 \, \text{Nm} \ + 3,600 \, \text{Nm} \]Solving for \(F'_f\): \[ F'_f \times 3.00 \, \text{m} = 14,380 \, \text{Nm} \] \[ F'_f = \frac{14,380 \, \text{Nm}}{3.00 \, \text{m}} = 4,793.33 \, \text{N} \]
04

Calculate Revised Rear Wheel Load

The new total weight on the truck is \(19,600 \, \text{N} + 3,600 \, \text{N} = 23,200 \, \text{N} \). The rear wheels' load \(F'_r\) is the remainder after accounting for the new front load: \[ F'_r = 23,200 \, \text{N} - 4,793.33 \, \text{N} = 18,406.67 \, \text{N} \]
05

Calculate the Weight to Lift the Front Wheels

For the front wheels to come off the ground, the torque around the rear axle must cause the normal force on the front wheels to drop to zero. This requires: \[ 10,780 \, \text{Nm} + W_{box} \times 1.00 \, \text{m} = F_r \times 3.00 \, \text{m} = 8,820 \, \text{N} \times 3.00 \, \text{m} \] Solving for \(W_{box}\): \[ 10,780 \, \text{Nm} + W_{box} \times 1.00 \, \text{m} = 26,460 \, \text{Nm} \] \[ W_{box} = 15,680 \, \text{N} \]

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Key Concepts

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

Equilibrium in physics
Equilibrium is a fundamental concept in physics that describes a state in which all forces and torques acting on a system cancel out, resulting in no net motion. In terms of a truck, as in our original exercise, equilibrium ensures that the truck remains stationary, even when additional weight is added or removed. When we talk about torque, which is the rotational equivalent of force, it must be balanced around a pivot point, which, in this scenario, is the rear axle of the truck.

To achieve equilibrium, the sum of the torques acting on the truck around any point must equal zero. This is critical because if one side experiences more torque, the truck could tip or shift, disrupting stability. This principle is applied in the exercise when it calculates how a box added to the truck tailgate influences equilibrium, requiring an adjustment in wheel loads. Thus, equilibrium ensures that the truck remains balanced and does not tilt or lift off the ground when loads change.
Load distribution
Load distribution in physics refers to how weight is spread across a structure or object. For vehicles like trucks, understanding load distribution involves looking at how weight from the truck body and any additional cargo is shared between the front and rear axles. This is crucial for maintaining balance and stability, especially during movement.

In the exercise, the total weight initially resting on the truck is divided between the front and the rear wheels. When the box is placed on the tailgate, it alters this distribution because additional weight introduces another force that shifts the balance of the vehicle. Calculating the new forces on each set of wheels after adding this extra weight requires understanding how to redistribute the loads effectively to maintain equilibrium.

The goal is to ensure that the additional load does not exceed the capacity of the axle or cause uneven wear and tear on the vehicle's components, ensuring safe and efficient operation.
Center of gravity
The center of gravity is a point where the entire weight of an object appears to act, regardless of its orientation in space. In physics, identifying the center of gravity is crucial for analyzing stability and balance, particularly in structures like vehicles.

For the truck in the exercise, its center of gravity helps determine how the weight of the truck and any additional load (like the box) affect its equilibrium. If the center of gravity is too far from the wheels' base support, it could lead to tipping or imbalance, particularly when the external loads shift.

When the box is added to the tailgate, it potentially shifts the truck's center of gravity towards the rear. Calculating how this new configuration interacts with the original balance of the truck is essential for ensuring that it does not tip but remains securely grounded. By calculating where the center of gravity falls relative to the wheelbase, one can preemptively address any issues with balanced load distribution and ensure the vehicle maintains optimal stability.

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Most popular questions from this chapter

A nonuniform beam 4.50 \(\mathrm{m}\) long and weighing 1.00 \(\mathrm{kN}\) makes an angle of \(25.0^{\circ}\) below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 3.00 \(\mathrm{m}\) farther down the beam and perpendicular to it (Fig. 11.31\()\) . The center of gravity of the beam is 2.00 \(\mathrm{m}\) down the beam from the pivot. Lighting equipment exerts a \(5.00-\mathrm{kN}\) downward force on the lower left endof the beam. Find the tension \(T\) in the cable and the horizontal and vertical components of the force exerted on the beam by the pivot. Start by sketching a free-body diagram of the beam.

Flying Buttress. (a) A symmetric building has a roof sloping upward at \(35.0^{\circ}\) above the horizontal on each side. If each side of the uniform roof weighs \(10,000 \mathrm{N}\) , find the horizontal force that this roof exerts at the top of the wall, which tends to push out the walls. Which type of building would be more in danger of collapsing: one with tall walls or one with short walls? Explain. (b) As you saw in part (a), tall walls are in danger of collapsing from the weight of the roof. This problem plagued the ancient builders of large structures. A solution used in the great Gothic cathedrals during the 1200 s was the flying buttress, a stone support running between the walls and the ground that helped to hold in the walls. A Gothic church has a uniform roof weighing a total of \(20,000 \mathrm{N}\) and rising at \(40^{\circ}\) above the horizontal at each wall. The walls are 40 \(\mathrm{m}\) tall, and a flying buttress meets each wall 10 \(\mathrm{m}\) below the base of the roof. What horizontal force must this flying buttress apply to the wall?

A \(1.05-\mathrm{m}\) -long rod of negligible weight is supported at its ends by wires \(A\) and \(B\) of equal length (Fig. 11.62\() .\) The crosssectional arca of \(A\) is 2.00 \(\mathrm{mm}^{2}\) und that of \(B\) is 4.00 \(\mathrm{mm}^{2} .\) Young's modulus for wire \(A\) is \(1.80 \times 10^{11} \mathrm{Pa}\) ; that for \(B\) is \(1.20 \times 10^{11} \mathrm{Pa}\) At what point along the rod should a weight \(w\) be suspended to produce (a) equal stresses in \(A\) and \(B,\) and \((b)\) equal strains in \(A\) and \(B ?\)

Prior to being placed in its hole, a \(5700-\mathrm{N}, 9.0\) -m-long, unjform utility pole makes some nonzero angle with the vertical. A vertical cable attached 2.0 \(\mathrm{m}\) below its upperend holds it in place while its lower end rests on the ground. (a) Find the tension in the cable and the magnitude and direction of the force exerted by the ground on the pole. (b) Why don't we need to know the angle the pole makes with the vertical, as long as it is not zero?

A \(350-N,\) uniform, \(1.50-m\) bar is suspended horizontally by two vertical cables at each end. Cable \(A\) can support a maximum tension of 500.0 \(\mathrm{N}\) without breaking, and cable \(B\) can support up to 400.0 \(\mathrm{N}\) . You want to place a small weight on this bart up is the heaviest weight you can put on without breaking either cable, and (b) where should you put this weight?
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