Two blocks are connected by a string over a frictionless, massless pulley such that one is resti

step 1 Okay, let's take a look at the 3 -body diagram for each block. So the 3 -body diagram for the ha

Two blocks are connected by a string over a frictionless,...

Key Concepts

Tension in a Massless String and Frictionless Pulley

The assumption of a massless string and a frictionless pulley implies that the tension is uniform throughout the string. This simplification allows the same tension force to be applied to both blocks and is critical for setting up the coupled equations derived from Newton’s Second Law, ultimately aiding in the calculation of the system’s acceleration and the tension force.

Frictional Force

Frictional force, specifically kinetic friction in this problem, opposes the relative motion of the block along the surface of the inclined plane. It is calculated as the product of the coefficient of kinetic friction and the normal force. Correctly incorporating this force into the equations is essential for accurately determining the system’s acceleration.

Free-Body Diagram

A free-body diagram is a schematic representation used to illustrate all the forces acting on an object. In the problem, drawing free-body diagrams for each block helps in identifying the gravitational forces, normal forces, frictional forces, and tension, thereby simplifying the process of setting up the equations required to solve for the unknowns in the system.

Newton's Second Law

Newton's Second Law is the foundation for analyzing the motion of the blocks, stating that the net force acting on an object equals its mass multiplied by its acceleration. In this context, it is applied separately to each block, leading to equations that incorporate gravitational force, friction, and tension to determine both the acceleration of the system and the tension in the connecting string.

Force Decomposition on an Inclined Plane

When dealing with inclined planes, the gravitational force acting on the block must be decomposed into two components: one parallel to the incline, which drives the block’s motion, and one perpendicular to the incline, which influences the normal force. Understanding this breakdown is crucial for accurately calculating the net force acting on the block on the inclined plane.

Transcript

00:01 Okay, let's take a look at the 3 -body diagram for each block.

00:04 So the 3 -body diagram for the hanging block is something like this, which means that the hanging block will experience a gravity from the earth, and the gravity is equal to mhg.

00:16 Mh is a mass of a hanging block.

00:18 And it will also experience the tension that was going upward, that the direction of a net force on the hanging block should be going downward.

00:26 It's because the hanging block is trying to move down.

00:31 Okay.

00:34 And the free body diagram for the block on an incline plane should be something like this.

00:40 And the angle between the incline plane and the ground is 45 degree.

00:44 And the block itself will experience a gravity from the earth, which is equal to mig.

00:51 Mi is the mass of a block from an incline plane.

00:55 And it experienced a tension that was moving upward, but along the direction of the incline plan.

01:03 And on the vertical direction, it will experience a normal force that was moving upward, which is perpendicular to the incline plane.

01:17 And lastly, you will also experience a friction that was along the incline plan, but in the opposite direction when it compares with the direction of the tension.

01:29 And the net force of the block on the incline plan should be moving upward, but along the incline plane.

01:38 Okay? so if we consider the upward direction on the incline plane is the positive x -axis, and the direction that is perpendicular to the incline plane's in an upward direction is the y -axis.

01:56 So therefore, we can decompose the gravity, which will have the x component of the gravity and the y component of the gravity.

02:03 Okay? and as you can tell, normal force is equal to the gravity times cosine 45 degree, which is equal to mig cosine 45 degree.

02:35 And we also know that friction in this case since it's kinetic friction, so it's equal to the coefficient of kinetic friction times the normal force, which is equal to mu times myg, cosine, cosine, 45 degree and the net force on the vertical direction that was perpendicular to the incline planes is equal to zero is because the block didn't moving up or down it only move along the incline plan okay so therefore we can set out the net force equation for the block on the incline plan which is f net i is equal to the tension minus the friction and then minus the x component of the gravity okay which is going which is also going to the opposite direction when you compare with the direction of the tension which is g sign 45 degree which is equal to this equation we go to t minus mu n minus mig sine 45 degree if we expand it we have t minus mu mig cosine 45 degree minus mig, sine 45 degree.

04:19 And we also know that the net force of the block on the incline plan is equal to the mass of the block on the incline plan times the acceleration a.

04:31 And this will equal to t minus mu, mu, mig, mig, sign 45 degree.

04:53 So it seems like we don't know how much tension it is.

04:56 But remember, the block on an incline plane was connected with the hanging block through a string.

05:04 Okay? which means that if we take a look at the free -by diagram and find the hanging block, we can have the net force equation for the hanging block.

05:12 I say f -net -h.

05:14 We use a green pencil...

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