A block with mass m1 is placed on an inclined plane with slope angle αand is connected to a hang

A block with mass m1 is placed on an inclined plane with...

A block with mass $m_1$ is placed on an inclined plane with slope angle $\alpha$ and is connected to a hanging block with mass $m_2$ by a cord passing over a small, frictionless pulley (Fig. P5.74). The coefficient of static friction is $\mu_s$, and the coefficient of kinetic friction is $\mu_k$. (a) Find the value of $m_2$ for which the block of mass $m_1$ moves up the plane at constant speed once it is set in motion. (b) Find the value of $m_2$ for which the block of mass $m_1$ moves down the plane at constant speed once it is set in motion. (c) For what range of values of $m_2$ will the blocks remain at rest if they are released from rest?

Key Concepts

Equilibrium Conditions and Newton's Second Law

For a body moving at constant speed, there is no net acceleration; therefore, the sum of forces acting on the system in any particular direction must be zero. This equilibrium condition allows us to set up force balance equations, which are essential for determining unknown quantities (like the mass m? in the examples) in systems involving friction and inclined planes.

Pulley Systems in Connected Bodies

In problems involving connected bodies with a pulley, the tension in the cord and the gravitational forces acting on the masses are interrelated. A frictionless pulley ensures that the tension is the same throughout the cord, simplifying the analysis of the system by allowing the use of common force equations for both bodies. This concept is essential for understanding how forces are transmitted in mechanical systems.

Force Decomposition on an Inclined Plane

This concept involves breaking down the gravitational force acting on the block into components that are parallel and perpendicular to the plane. The parallel component (mg sin ?) influences the motion along the plane, while the perpendicular component (mg cos ?) determines the normal force, which is crucial for calculating frictional forces.

Frictional Forces (Static and Kinetic Friction)

Friction plays a key role in resisting motion along surfaces. Static friction is the force that must be overcome to initiate motion and has a variable magnitude up to a maximum value (?s times the normal force). Once motion begins, kinetic friction takes over, which usually has a constant magnitude (?k times the normal force). Understanding the distinction and application of these coefficients is vital in problems involving the transition from rest to motion.

Transcript

00:01 In this problem, we have block one on an incline plane of angle alpha, and it's tethered to block two, which is hanging over the edge.

00:11 If we draw the free body diagram for block one, we have the tension from rope to pulling it up the ramp this way.

00:18 Our normal force is normal to the ramp surface, so it's this way.

00:22 The force of friction is opposing the motion, so if block one is going up the ramp, then the force of friction is pushing this way.

00:29 And then we also have our force of gravity downward.

00:33 Since the angle of our ramp is alpha, we know this angle is alpha.

00:37 And that means by trigonometry, this must be equal to mg sine alpha.

00:45 And this side must be equal to mg cosine times a cosine of alpha.

00:51 Now to solve part a, we need to notice that even when block 1 is moving at constant velocity, its acceleration is zero.

00:59 And so it's net force is zero.

01:01 So we have the sum of the forces is equal to zero.

01:06 And since it's moving in this direction, we know the sum of the forces in that direction is equal to zero.

01:11 So we need the two is decompose all our forces into the ramp direction, which i'll just call the x direction.

01:19 And then this can be the y direction.

01:22 And so we need the x components of all our forces.

01:25 We're going to sum them up and set them equal to zero.

01:27 That's the condition for constant velocity block sliding in this direction.

01:31 So we know that the tension points in the positive direction, but the force of friction points in the negative direction.

01:38 Also, the contribution from the normal force will point in the negative direction.

01:48 Oh, actually, sorry, i just got confused.

01:51 The normal force is in the y direction only.

01:53 It's the x component of the gravity force, which we'll be considering.

01:58 And so when we set this up, we will have, this implies negative force of friction.

02:04 I'm pulling the negative out of the force of friction.

02:06 Because i already know it must be negative.

02:09 So when we calculate this, we have to remember not to make it negative again.

02:14 And we're going to subtract m1 times gravity times sign of alpha, which is the component here from the gravity.

02:24 And then we're going to add in the force of tension, which is purely the force here, which is m2 times g...

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