An 8 -kg cylinder C rests on a 4 -kg platform A supported by...

Key Concepts

Normal Force and Reaction Forces

The normal force is a contact force exerted by a surface to support the weight of an object resting upon it. In dynamics, analyzing how additional acceleration or applied forces modify the effective normal force helps in determining the actual force interactions between objects in contact, such as the force a cylinder exerts on a platform.

Pulley Systems and Tension

In systems involving pulleys, ropes transmit tension forces from one mass to another, and often constraints relate the motions of different parts of the system. Understanding these constraints and how the tension is distributed through the pulley system is essential in analyzing the motion and forces acting on connected objects.

Kinematics of Uniformly Accelerated Motion

Kinematics deals with the description of motion without considering its causes. In systems where acceleration is constant, kinematic equations are used to relate time, velocity, displacement, and acceleration. This concept is vital for calculating the velocity of an object after a given time interval of accelerated motion.

Newton's Second Law

Newton’s Second Law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. This principle underlies the analysis of forces acting on each component of a mechanical system and is fundamental in setting up equations to determine unknown accelerations and tensions.

Free-body Diagrams

Free-body diagrams are graphical representations that isolate an object and depict all external forces acting upon it. They help in visualizing how forces such as gravitational, tension, and normal forces interact, and are crucial in applying Newton’s laws to solve for unknown quantities in mechanical systems.

Transcript

00:01 In this problem, we have a system containing of a cylinder which is resting on a platform supported by a cord, passing over a pulley and attached to another block.

00:13 All the masses are known and the system is released from rest.

00:17 We want to calculate the velocity of block b after a given time and the force exerted by the cylinder on the platform.

00:25 So initially, if we look at the system of blocks a and c and we draw a momentary.

00:32 Impulse diagram we can see initially the system is at rest so a and c have combined momentum m a plus m c times v which is equal to zero when the system is released from rest there's an impulse acting along the cord which is the tension in the core t times the time t and the weight acting downward the weight is the mass of a plus the mass of c times g times time t.

01:02 And the final state of the system is when c and a move down together with momentum m .a plus m .c times its final velocity.

01:14 So if we look at blocks a and c mathematically, and we take the downward direction to be positive, we can write it as follows.

01:36 So, ma plus mc times the initial velocity v1 minus the tension times the time in the opposite direction, plus the weight times time, which is the mass, ma plus mc times g, times the same time t, during which the impulse acts, must equal to the final momentum of the system, which is ma plus mc times v, 2, the second state of the system.

02:20 So we know initially the system is addressed.

02:25 So 0 plus the forces 12g minus t any time of 0 .8 seconds, so little t is 0 .8 must equal to the final momentum of the system which is 12 times v so we have two equations and two unknowns in this equation so we'll call it equation one now let's do the same for block a for block b alone we'll look at block b and again we'll take the downward direction as positive actually for block b we'll take the upward direction as positive block a moves in the opposite direction so a block b so for block b alone the initial momentum of block which is its mass mb times v1 plus the tension t time t which acts in the positive direction upward minus its weight m b g times time t so the two impulses due to the tension and the weight and this must equal to the final momentum of block b mb v2 now remember number v is the same for the system of block a and c and for b as well, they will travel at the same velocity.

04:05 So if we rearrange this equation, we know that initially the system is addressed, so there's no momentum.

04:11 That's 0 plus t minus 4g into a time of 0 .8 seconds that's equal to 4v.

04:25 And again, we have the same two unknowns.

04:27 The tension and the speed at which the block is moving...

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