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$\bullet$ A uniform solid cylinder of mass $M$ is supported on a ramp that rises at an angle $\theta$ above the horizontal by a wire that is wrapped around its rim and pulls on it tangentially parallel to the ramp (Fig. 10.79$)$ . (a) Show that there must be friction on the surface for the cylinder to balance this way. (b) Show that the tension in the wire must be equal to the friction force, and find this tension.

a) See drawingb) $\frac{1}{2} M g \sin \theta$

Physics 101 Mechanics

Chapter 10

Dynamics of Rotational Motion

Newton's Laws of Motion

Rotation of Rigid Bodies

Equilibrium and Elasticity

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Hope College

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Hello. We're working Problem 69 which is a uniformed solid cylinder that is of mass M that is on an incline plane at an angle of data. The cylinder is being pulled up this ramp by a tension force. I'm, um we know it has a weight of mg. You know, there's a normal force at the contact point perpendicular up, and it's ask. The first question it asks is, How do you know that there must be a friction force opposing motion. So if it opposes motion, we know this friction force is going this way. How do we know that? How do we know that exists? Because, um, we know that this is not rotating up the hill. So we know that net torque must be zero now, which forces are going to provide to work well or weight will not because it's through the center. Our normal force will not provide to work because it goes. It has no level arm because it travels through the center. We know tension will provide torque based on the radius, and then in order for it to be zero, there must be some other force there, and that is In fact, friction opposes it in provides an opposite torque at the same radius. So we know that the tension and friction must be equal to each other in order for it to stay balanced, Our second question shows says, show that the tension equals a fiction force and then find that tension. So there we showed that the tension force equals the friction force. Now let's find the tension. So we find it a couple of ways here. We know that are looking at our net forces in the X direction. Now, remember this angle stada. So we know in the ex direction that it must be equal to zero. So we know that tension plus friction must equal this component this horizontal component of the weight so mg sign of Seita that so we already just verified in fact, that tension and friction are the same so we can sub in friction For attention. We get Angie, sign data that's solving four friction becomes too f here. So our friction force actually mg signed data divided by two Thanks

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