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The tires of a \(1500-\mathrm{kg}\) car are \(0.600 \mathrm{m}\) in diameter, and the coefficients of friction with the road surface are \(\mu_{s}=0.800\) and \(\mu_{k}=0.600 .\) Assuming that the weight is evenly distributed on the four wheels, calculate the maximum torque that can be exerted by the engine on a driving wheel without spinning the wheel. If you wish, you may assume the car is at rest.

Short Answer

Expert verified
The maximum torque that can be exerted by the engine on a driving wheel without spinning it is \(882 N*m.\)

Step by step solution

01

Calculate the weight of the car

First, calculate the weight (W) of the car. Weight, which acts downwards due to the force of gravity, is equal to the mass (m) of the object times the acceleration due to gravity (g, approximately \(9.8 m/s^{2}\) on the surface of the Earth). Therefore, the weight of the car \(W = m*g = 1500 kg * 9.8 m/s^{2} = 14700 N.\)
02

Calculate the normal force per wheel

Next, calculate the normal force (N) on each wheel. Since the weight is evenly distributed on the four wheels, the normal force on each wheel is a quarter of the total weight. Therefore, \(N = W / 4 = 14700 N / 4 = 3675 N.\)
03

Calculate the static friction force

Now, calculate the maximum value of the static friction (fs) which equals the coefficient of static friction (μs) times the normal force (N). Therefore, the static friction is \(fs = μs * N = 0.800 * 3675 N = 2940 N.\)
04

Calculate the torque

Finally, calculate the maximum torque that can be exerted by the engine without spinning the wheel. Torque (Ï„) equals force (F) times distance (d), with force being the static friction and distance being the radius (r) of the wheel. Therefore, the torque is \(Ï„ = fs * r = 2940 N * 0.300 m = 882 N*m.\) This is the maximum torque that can be exerted by the engine on a driving wheel without spinning it.

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Key Concepts

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

Static Friction Force
When a car attempts to accelerate, its wheels exert a force against the road due to static friction. This force is crucial as it allows the vehicle to move forward without slipping. The static friction force is the maximum force that can be exerted before the wheel begins to slip. In scientific terms, static friction arises when two surfaces are in contact but not moving relative to each other. In our example, the static friction force is computed using the coefficient of static friction (usually denoted as \( \mu_s \) ) and the normal force - the perpendicular force exerted by the surface. The formula for the maximum static friction force is \( f_s = \mu_s \times N \). This value represents the threshold above which the tires will begin to skid. It's a key factor in determining the maximum torque applicable to a car's wheel for acceleration without spinning.

A higher coefficient of static friction means the tires can tolerate more torque before slipping. On the contrary, icy or wet conditions can significantly reduce static friction, increasing the chance of wheel spin. Understanding static friction is not only essential for driver safety but also for designing vehicles that maximize traction and performance.
Normal Force Calculation
Normal force is the support force exerted upon an object that is in contact with another stable object. Imagine holding a book against a table; the table exerts an upward force equal to the weight of the book—this is the normal force. For vehicles, each tire experiences a normal force from the road, which is a reaction to the vehicle's weight due to gravity. In the case of our hypothetical car, which weighs 1500 kilograms, this normal force is what allows the calculation of static friction force.

The formula to determine the normal force per wheel is as simple as dividing the car's total weight by the number of wheels, which, in most cars, is four. Mathematically, it is expressed as \( N = \frac{W}{4} \), where \(W\) is the weight of the car. Normal force plays a pivotal role in driving dynamics because it directly impacts traction, braking, and cornering capabilities. By calculating the normal force, engineers can optimize tire and suspension design to improve overall vehicle performance and safety.
Torque and Wheel Radius
Torque is a measure of how much a force acting on an object causes that object to rotate. In the context of vehicles, the engine generates torque that rotates the wheels and propels the car forward. The torque transferred to the wheel can be maximized up to the point where static friction is overcome, resulting in wheel spin. To calculate the torque that leads to this threshold, we use the formula \( \tau = f_s \times r \), where \( \tau \) is the torque, \( f_s \) is the static friction force, and \( r \) is the radius of the wheel.

The radius of the wheel is a crucial factor; it determines the leverage force applied. A larger radius will require more force to produce the same amount of torque. Therefore, if two cars exert the same amount of force, but one has larger wheels, it will generate less torque. By understanding the relationship between torque and wheel radius, automotive engineers can determine the optimum wheel size for a vehicle, balancing performance and efficiency to meet specific driving requirements. The importance of torque in vehicle dynamics can never be overstated as it impacts acceleration, fuel economy, and overall driving performance.

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Most popular questions from this chapter

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A thin rod of mass \(0.630 \mathrm{kg}\) and length \(1.24 \mathrm{m}\) is at rest, hanging vertically from a strong fixed hinge at its top end. Suddenly a horizontal impulsive force \((14.7 \mathrm{i}) \mathrm{N}\) is applied to it. (a) Suppose the force acts at the bottom end of the rod. Find the acceleration of its center of mass and the horizontal force the hinge exerts. (b) Suppose the force acts at the midpoint of the rod. Find the acceleration of this point and the horizontal hinge reaction. (c) Where can the impulse be applied so that the hinge will exert no horizontal force? This point is called the center of percussion.

A car accelerates uniformly from rest and reaches a speed of \(22.0 \mathrm{m} / \mathrm{s}\) in \(9.00 \mathrm{s} .\) If the diameter of a tire is \(58.0 \mathrm{cm}\) find (a) the number of revolutions the tire makes during this motion, assuming that no slipping occurs. (b) What is the final angular speed of a tire in revolutions per second?

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