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Chapter 8: Problem 109

A bicycle travels up an incline at constant velocity. The magnitude of the frictional force due to the road on the rear wheel is \(f=3.8 \mathrm{N} .\) The upper section of chain pulls on the sprocket wheel, which is attached to the rear wheel, with a force \(\overrightarrow{\mathbf{F}}_{\mathrm{C}} .\) The lower section of chain is slack. If the radius of the rear wheel is 6.0 times the radius of the sprocket wheel, what is the magnitude of the force \(\overrightarrow{\mathbf{F}}_{\mathrm{C}}\) with which the chain pulls?

Short Answer

Expert verified
Answer: The magnitude of the force exerted by the chain on the sprocket wheel is 22.8 N.

Step by step solution

01

Identify the torques acting on the rear wheel

To analyze the torques acting on the rear wheel, we need to consider the forces involved: frictional force (f) due to the road, and the force exerted by the chain (\(\overrightarrow{\mathbf{F}}_{\mathrm{C}}\)).
02

Calculate the torque due to frictional force

The torque due to frictional force acting on the rear wheel can be calculated using the following formula: $$\tau_{f} = f \times r_{\mathrm{R}}$$ where \(\tau_{f}\) is the torque due to the frictional force, \(f\) is the frictional force, and \(r_{\mathrm{R}}\) is the radius of the rear wheel.
03

Calculate the torque due to chain force

The torque due to the chain force acting on the sprocket wheel can be calculated using the following formula: $$\tau_{\mathrm{C}} = F_{\mathrm{C}} \times r_{\mathrm{S}}$$ where \(\tau_{\mathrm{C}}\) is the torque due to the chain force, \(F_{\mathrm{C}}\) is the force exerted by the chain, and \(r_{\mathrm{S}}\) is the radius of the sprocket wheel.
04

Relate the radii of the rear wheel and the sprocket wheel

According to the problem, the radius of the rear wheel is 6 times the radius of the sprocket wheel. So, we can write this relation as: $$r_{\mathrm{R}} = 6r_{\mathrm{S}}$$
05

Equate the torques to find the chain force

Since the bicycle is moving at a constant velocity, the net torque should be zero. Therefore, we can equate the torques due to the chain force and the frictional force as follows: $$\tau_{\mathrm{C}} = \tau_{f}$$ Substituting the expressions for the torques and the relation between the radii into this equation, we get: $$F_{\mathrm{C}}\times r_{\mathrm{S}} = f \times r_{\mathrm{R}}$$ Now, substituting the relation between the radii, we have: $$F_{\mathrm{C}} \times r_{\mathrm{S}} = f \times (6r_{\mathrm{S}})$$ Dividing both sides by \(r_{\mathrm{S}}\), we get: $$F_{\mathrm{C}}= 6f$$
06

Calculate the magnitude of the chain force

Now, we can substitute the given value of frictional force \(f = 3.8\mathrm{N}\) into the equation to find the magnitude of the chain force: $$F_{\mathrm{C}} = 6\times 3.8\mathrm{N}$$ $$F_{\mathrm{C}} = 22.8\mathrm{N}$$ Thus, the magnitude of the force exerted by the chain on the sprocket wheel is \(22.8\mathrm{N}\).

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Most popular questions from this chapter

Two children standing on opposite sides of a merry-goround (see Fig. 8.5 ) are trying to rotate it. They each push in opposite directions with forces of magnitude \(10.0 \mathrm{N} .\) (a) If the merry-go-round has a mass of $180 \mathrm{kg}\( and a radius of \)2.0 \mathrm{m},$ what is the angular acceleration of the merry-go-round? (Assume the merry-go-round is a uniform disk.) (b) How fast is the merry-go-round rotating after \(4.0 \mathrm{s} ?\)
Derive the rotational form of Newton's second law as follows. Consider a rigid object that consists of a large number \(N\) of particles. Let \(F_{i}, m_{i},\) and \(r_{i}\) represent the tangential component of the net force acting on the ith particle, the mass of that particle, and the particle's distance from the axis of rotation, respectively. (a) Use Newton's second law to find \(a_{i}\), the particle's tangential acceleration. (b) Find the torque acting on this particle. (c) Replace \(a_{i}\) with an equivalent expression in terms of the angular acceleration \(\alpha\) (d) Sum the torques due to all the particles and show that $$\sum_{i=1}^{N} \tau_{i}=I \alpha$$
A Ferris wheel rotates because a motor exerts a torque on the wheel. The radius of the London Eye, a huge observation wheel on the banks of the Thames, is \(67.5 \mathrm{m}\) and its mass is \(1.90 \times 10^{6} \mathrm{kg} .\) The cruising angular speed of the wheel is $3.50 \times 10^{-3} \mathrm{rad} / \mathrm{s} .$ (a) How much work does the motor need to do to bring the stationary wheel up to cruising speed? [Hint: Treat the wheel as a hoop.] (b) What is the torque (assumed constant) the motor needs to provide to the wheel if it takes 20.0 s to reach the cruising angular speed?
A skater is initially spinning at a rate of 10.0 rad/s with a rotational inertia of \(2.50 \mathrm{kg} \cdot \mathrm{m}^{2}\) when her arms are extended. What is her angular velocity after she pulls her arms in and reduces her rotational inertia to \(1.60 \mathrm{kg} \cdot \mathrm{m}^{2} ?\)
A person is doing leg lifts with 3.0 -kg ankle weights. She is sitting in a chair with her legs bent at a right angle initially. The quadriceps muscles are attached to the patella via a tendon; the patella is connected to the tibia by the patella tendon, which attaches to bone \(10.0 \mathrm{cm}\) below the knee joint. Assume that the tendon pulls at an angle of \(20.0^{\circ}\) with respect to the lower leg, regardless of the position of the lower leg. The lower leg has a mass of \(5.0 \mathrm{kg}\) and its center of gravity is $22 \mathrm{cm}\( below the knee. The ankle weight is \)41 \mathrm{cm}$ from the knee. If the person lifts one leg, find the force exerted by the patella tendon to hold the leg at an angle of (a) \(30.0^{\circ}\) and (b) \(90.0^{\circ}\) with respect to the vertical.
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