/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 A pickup truck has a wheelbase o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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) A box weighing 3600 \(\mathrm{N}\) is now placed on the tailgate, 1.00 \(\mathrm{m}\) behind the rear axle. 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 tailgate to make the front wheels come off the ground?

Short Answer

Expert verified
Front wheels support 8320 N, rear wheels support 11,280 N. 8085 N is needed on the tailgate to lift the front wheels.

Step by step solution

01

Calculating the Initial Weight Distribution

Initially, the truck's total weight is distributed as 10,780 N on the front wheels and 8820 N on the rear wheels. This distribution on the level road means the total weight supported by both axles is 19,600 N (10,780 N + 8820 N).
02

Understanding Torque and Lever Arm

When a box weighing 3600 N is placed on the tailgate, it applies a torque about the rear axle. This will affect the weight distribution between the front and rear wheels. The lever arm for this torque is 1.00 m because it is placed 1.00 m behind the rear axle.
03

Calculating the Effect of the Box on the Rear Wheels

Because the box is placed behind the rear axle, we need to calculate the additional weight supported by the rear wheels: the torque exerted on the rear wheels makes the rear wheels support its entire weight of 3600 N, plus some of the initially balanced weight from the front wheels.
04

Calculating Total Weight on the Front Wheels

To find out how much weight now rests on the front wheels, we need to consider the effect of torque shifting weight from the front wheels to the rear: \[ (10,780 \text{ N} \times 3.00 \text{ m}) - (3,600 \text{ N} \times 4.00 \text{ m}) = W_f \times 3.00 \text{ m} \] Solving this gives us the new force on the front wheels \(W_f\). After simplification:\[W_f = \frac{(10,780 \text{ N} \times 3.00 \text{ m}) - (3,600 \text{ N} \times 4.00 \text{ m})}{3.00 \text{ m}} = 8320 \text{ N} \] So, the front wheels now support 8320 N.
05

Calculating Total Weight on the Rear Wheels

The weight on the rear wheels is now the initial rear weight plus the weight of the box plus the weight removed from the front wheels:\[W_r = 19,600 \text{ N} - 8320 \text{ N} = 11,280 \text{ N} \]Now, the rear wheels support 11,280 N.
06

Calculating Needed Weight to Lift Front Wheels

To make the front wheels come off the ground, the entire torque must shift entirely to the rear wheels. This requires:\[ 10,780 \text{ N} \times 3.00 \text{ m} = W \times 4.00 \text{ m} \] Solving for \(W\): \[ W = \frac{10,780 \text{ N} \times 3.00 \text{ m}}{4.00 \text{ m}} = 8085 \text{ N} \] So, 8085 N would need to be placed on the tailgate to lift the front wheels.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Torque and Lever Arm
When the box is placed on the tailgate of the truck, it creates torque around the rear axle. But what exactly is torque? Torque is a measure of the tendency of a force to rotate an object about an axis. In simpler terms, torque is what causes an object to spin or turn.
The formula for torque \(\tau\) is given by: \(\tau = F \times d\), where \(F\) is the force applied, and \(d\) is the distance from the axis of rotation to where the force is applied, known as the lever arm.
In this problem, the box adds a force of 3600 N at a distance (lever arm) of 1.00 m behind the rear axle. This torque affects how the weight is distributed between the front and rear wheels. The placement of the box shifts the vehicle's balance, demonstrating the power of torque to change how loads are supported.
Weight Distribution
Weight distribution is crucial in determining how the truck remains balanced. Initially, the truck's weight is divided between the front and rear wheels with 10,780 N resting on the front wheels and 8820 N on the rear.
When the box is placed on the tailgate, it not only adds weight to the truck but also changes the distribution of that weight. This shift occurs because the added mass at the rear increases the torque around the rear axle. The result is a redistribution of weight, decreasing the load on the front wheels.
If more weight is needed on the rear to lift the front wheels off the ground, we should add enough mass to counteract the torque created by the entire weight previously on the front wheels. This way, the balance is tipped entirely to the back, illustrating how placement of weight affects stability.
Newton's Laws of Motion
The behavior of the truck and the effect of placing a weight on the tailgate can be understood clearly through Newton's Laws of Motion. Newton's First Law states that an object will remain at rest or in uniform motion unless acted upon by a net external force. In the truck's case, it stays static until forces (like the added weight and resulting torque) act upon it.
Newton's Second Law explains how the acceleration (or in this case, the shift in load) occurs when the mass of the box changes the equilibrium of forces acting on the truck. The mass, combined with gravity, introduces a force that shifts the weight distribution.
Finally, Newton's Third Law says that for every action, there is an equal and opposite reaction. Perhaps less obvious in this context, but significant nonetheless: the road exerts an upward force on the truck, counteracting the forces due to the truck's weight, and this interplay keeps the truck in balance. Understanding these laws helps in analyzing how weight placement and resultant forces affect movement and balance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A uniform rod is 2.00 \(\mathrm{m}\) long and has 1.80 \(\mathrm{kg} . \mathrm{A}\) \(2.40-\) -kg clamp is attached to the rod. How far should the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be 1.20 \(\mathrm{m}\) from the left-hand end of the rod?

Cp A steel cable with cross-sectional area 3.00 \(\mathrm{cm}^{2}\) has an elastic limit of \(2.40 \times 10^{8}\) Pa. Find the maximum upward acceleration that can be given a 1200 -kg elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.

CP A uniform drawbridge must be held at a \(37^{\circ}\) angle above the horizontal to allow ships to pass underneath. The drawbridge weighs \(45,000 \mathrm{N}\) and is 14.0 \(\mathrm{m}\) long. A cable is connected 3.5 \(\mathrm{m}\) from the hinge where the bridge pivots (measured along the bridge \()\) and pulls horizontally on the bridge to hold it in place. (a) What is the tension in the cable? (b) Find the magnitude and direction of the force the hinge exerts on the bridge. (c) If the cable suddenly breaks, what is the magnitude of the angular acceleration of the angular speed of the drawbridge as it breaks? (d) What is the angular speed of the drawbridge as it becomes horizontal?

A bookcase weighing 1500 \(\mathrm{N}\) rests on a horizon- tal surface for which the coefficient of static friction is \(\mu_{\mathrm{s}}=0.40 .\) The bookcase is 1.80 \(\mathrm{m}\) tall and 2.00 \(\mathrm{m}\) wide; its center of gravity is at its geo- metrical center. The bookcase rests on four short legs that are each 0.10 \(\mathrm{m}\) from the edge of the bookcase. A person pulls on a rope attached to an upper corner of the bookcase with a force \(\vec{\boldsymbol{F}}\) that makes an angle \(\theta\) with the bookcase (Fig. \(P 11.97 )\) . (a) If \(\theta=90^{\circ},\) so \(\vec{F}\) is horizontal, show that as \(F\) is increased from zero, the bookcase will start to slide before it tips, and calculate the magnitude of \(\vec{\boldsymbol{F}}\) that will start the bookcase sliding. (b) If \(\theta=0^{\circ},\) so \(\vec{\boldsymbol{F}}\) is vertical, show that the bookcase will tip over rather than slide, and calculate the magnitude of \(\vec{\boldsymbol{F}}\) that will cause the bookcase to start to tip. (c) Calculate as a function of \(\theta\) the magnitude of \(\vec{\boldsymbol{F}}\) that will cause the bookcase to start to slide and the magnitude that will cause it to start to tip. What is the smallest value that \(\theta\) can have so that the bookcase will still start to slide before it starts to tip?

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 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?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.