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The wheel of a car has a radius of 0.350 m. The engine of the car applies a torque of 295 N m to this wheel, which does not slip against the road surface. Since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a countertorque. Moreover, the car has a constant velocity, so this countertorque balances the applied torque. What is the magnitude of the static frictional force?

Short Answer

Expert verified
The magnitude of the static frictional force is approximately 843 N.

Step by step solution

01

Understand the Relationship between Torque and Force

Torque (\( \tau \)) is the product of the force (\( F \)) and the lever arm's length (distance from the pivot point, which is the radius of the wheel in this case, \( r \)). The formula for torque is:\[\tau = F \times r\]
02

Use Given Values for Torque and Radius

We are given the torque applied by the engine as 295 N m, and the radius of the wheel is 0.350 m. We need to solve for the static frictional force (\( F_s \)).
03

Rearrange Torque Formula to Solve for Static Frictional Force

Rearrange the torque formula to solve for the force:\[ F_s = \frac{\tau}{r} \]
04

Substitute Given Values into the Formula

Substitute the known values of torque and radius into the rearranged equation:\[ F_s = \frac{295 \, \text{N m}}{0.350 \, \text{m}} \]
05

Calculate the Magnitude of Static Frictional Force

Perform the calculation:\[ F_s = \frac{295}{0.350} \approx 842.857 \, \text{N} \]Rounding the result gives approximately 843 N.

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Key Concepts

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

Understanding Torque
Torque is a measure of how much a force acting on an object causes that object to rotate. It is the rotational equivalent of a linear force. Imagine you are trying to open a door; the further away from the hinge (pivot point) you push, the easier it is to rotate the door. This distance is known as the lever arm.

- Torque, often represented by the symbol \( \tau \), is calculated using the formula \( \tau = F \times r \), where \( F \) is the force applied, and \( r \) is the distance from the pivot point.
- The unit for torque is the Newton meter (N m), which combines the force unit (Newton) and distance unit (meter).

In many mechanical systems like a car, torque is critical. It represents the engine's power to rotate the wheels, which propels the vehicle forward. Remember, if there are no external forces to balance the applied torque, the object will start to rotate.
Exploring Car Wheel Dynamics
Car wheel dynamics involves how forces interact with the wheels to generate motion. When a car moves, the engine applies a torque to the wheels. This torque must be converted to linear motion for the car to drive forward.

- One essential force involved is static friction. This force is critical because it prevents the wheels from slipping against the road surface as they rotate.
- Essentially, static friction provides the grip necessary to ensure that torque leads to forward movement rather than skidding.

Static friction also creates a countertorque that opposes the engine's applied torque. For a car in constant velocity, these torques must balance, ensuring the vehicle maintains its speed without accelerating or decelerating. This balance of forces is crucial for smooth and controlled driving.
Maintaining Mechanical Equilibrium
When a system experiences no net force or torque, it is in a state of mechanical equilibrium. For our car example, achieving mechanical equilibrium ensures that the car maintains its constant velocity without the wheels slipping.

- In mechanical equilibrium, the sum of all forces and the sum of all torques in a system are zero.
- This balance allows the car to stay in control, as any imbalance would result in acceleration, deceleration, or spinning.

Static friction plays a crucial role in maintaining this balance. As the car moves forward, static friction generates a countertorque that perfectly balances the engine's torque, keeping the wheel rotating at a steady pace.

Understanding mechanical equilibrium helps in designing systems where stability and safety are paramount, like in vehicles and machinery, where control over motion is critical.

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Most popular questions from this chapter

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 300-mile trip in a typical midsize car produces about \(1.2 \times 10^{9} \mathrm{J}\) of energy. How fast would a \(13-\mathrm{kg}\) flywheel with a radius of 0.30 \(\mathrm{m}\) have to rotate to store this much energy? Give your answer in rev/min.

Two children hang by their hands from the same tree branch. The branch is straight, and grows out from the tree trunk at an angle of 27.0 above the horizontal. One child, with a mass of 44.0 kg, is hanging 1.30 m along the branch from the tree trunk. The other child, with a mass of 35.0 kg, is hanging 2.10 m from the tree trunk. What is the magnitude of the net torque exerted on the branch by the children? Assume that the axis is located where the branch joins the tree trunk and is perpendicular to the plane formed by the branch and the trunk.

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has a length of 0.80 m and is uniform. It is hanging vertically straight downward. The end of the rod nearest the floor is given a linear speed \(v_{0},\) so that the rod begins to rotate upward about the pivot. What must be the value of \(v_{0}\) . such that the rod comes to a momentary halt in a straight-up orientation, exactly opposite to its initial orientation?

A small 0.500-kg object moves on a frictionless horizontal table in a circular path of radius 1.00 m. The angular speed is 6.28 rad/s. The object is attached to a string of negligible mass that passes through a small hole in the table at the center of the circle. Someone under the table begins to pull the string downward to make the circle smaller. If the string will tolerate a tension of no more than 105 N, what is the radius of the smallest possible circle on which the object can move?

A 15.0-m length of hose is wound around a reel, which is initially at rest. The moment of inertia of the reel is \(0.44 \mathrm{kg} \cdot \mathrm{m}^{2},\) and its radius is 0.160 \(\mathrm{m}\) . When the reel is turning, friction at the axle exerts a torque of magnitude 3.40 \(\mathrm{N} \cdot \mathrm{m}\) on the reel. If the hose is pulled so that the tension in it remains a constant \(25.0 \mathrm{N},\) how long does it take to completely unwind the hose from the reel? Neglect the mass and thickness of the hose on the reel, and assume that the hose unwinds without slipping.

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