A marble block of mass m1=559.1 kg and a granite block of mass m2=128.4 kg are connected to ea

step 1 Once again, welcome to a new problem. This time we are dealing with forces. And when you think a

A marble block of mass m1=559.1 kg and a granite block of...

A marble block of mass $m_{1}=559.1 \mathrm{~kg}$ and a granite block of mass $m_{2}=128.4 \mathrm{~kg}$ are connected to each other by a rope that runs over a pulley as shown in the figure. Both blocks are located on inclined planes with angles $\alpha=38.3^{\circ}$ and $\beta=57.2^{\circ} .$ The rope glides over the pulley without fric. tion, but the coefficient of friction between block 1 and the inclined plane is $\mu_{1}=0.13,$ and that between block 2 and the inclined plane is $\mu_{2}=0.31$. (For simplicity, assume that the coefficients of static and kinetic friction are the same in each case.) What is the acceleration of the marble block? Note that the positive $x$ -direction is indicated in the figure.

Key Concepts

Newton's Second Law

This principle states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. It is fundamental when analyzing the motion of objects, allowing the setup of equations that factor in gravitational, frictional, and tension forces acting on each object.

Force Decomposition on Inclined Planes

When objects are placed on inclined surfaces, gravitational forces are broken down into components parallel and perpendicular to the incline. The parallel component drives motion along the plane, while the perpendicular component determines the normal force, which is essential for calculating friction.

Friction on Inclined Planes

Friction, arising from the interaction between surfaces, opposes sliding motion. It is calculated using the coefficient of friction and the normal force, and it critically affects the net force and subsequent acceleration of objects on inclines.

Systems of Connected Bodies

In scenarios where bodies are linked by a rope over a pulley, their motions are interdependent. The tension in the rope and the common acceleration constraint allow for writing simultaneous equations for each body, which are solved to understand the overall system behavior.

Free Body Diagrams

This tool involves drawing each object separately and illustrating all forces acting on it. Free body diagrams help in visualizing and breaking down the forces such as gravity, friction, normal, and tension, thus aiding in the systematic application of Newton's laws.

Transcript

00:03 Once again, welcome to a new problem.

00:06 This time we are dealing with forces.

00:09 And when you think about newton's second law, what tends to happen is that force is the product of the mass and the acceleration.

00:22 So that's newton's second law.

00:25 And whenever there's equilibrium, and we're thinking about equilibrium in terms of, let's say you have a box that's sitting on the table.

00:33 And the box has a weight m .g and the weight is opposed by the reaction of the box.

00:44 So this is newton's third law where we say action and reaction forces, action and reaction forces equal and opposite.

01:00 So action and reaction forces are equal and opposite.

01:03 It.

01:05 So newton's third law helps us with equilibrium where we're saying the sum of forces in the x direction is zero and the sum of forces in the y direction is also zero and that tends to happen whenever there's equilibrium.

01:25 So in this new problem we have an incline.

01:34 Okay, we have an incline and it's shaped such that there is a block on both directions.

01:43 So this is our incline and on the incline there sits two boxes.

01:49 The first box sits right here and then the second box sits on this other side like that.

01:58 The first box has a mass m1 and then the other has a mass m2.

02:06 So looking at the problem, we see that the marble block, the marble block that's m1 has a mass equivalent to 559 .1.

02:24 So this is a marble block.

02:27 This is a marble block.

02:29 And then we have another marble block, m2.

02:32 Well, the second block, the second block, sorry.

02:37 The second block is a granite block and it has a mass m2 that's equivalent to 128 .4 kilograms.

02:49 So that's a granite block we're seeing on the left.

02:54 And then also the two blocks are connected with the rope that runs through a pulley system.

03:07 So there's a pulley right here.

03:10 This is a pulley.

03:13 And then what happens is that we're going to have a rope that runs around like this.

03:24 I want to go back a little bit.

03:26 So we have a rope that runs around like this.

03:31 And it connects these two.

03:36 So these two are connected by a rope, connected by a pulley.

03:43 That runs around like that.

03:47 And the inclines on two angles.

03:50 Of course, the first one is alpha, and alpha happens to be 38 .3 degrees.

04:00 And then the second one is beta.

04:02 And beta happens to be equivalent to 57 .2 degrees.

04:10 And then also the other requirement is that we're going to have friction.

04:18 There is no friction on the rope.

04:22 So there's no friction on the row.

04:26 We don't have any friction over there.

04:30 But then block one deals with friction.

04:34 So block one deals with friction and the coefficient of friction for block one.

04:46 The coefficient of friction for block one is equivalent to, 0 .13 and then the coefficient of friction for block 2 is the same as is the same as 0 .31 the coefficient of friction we have for block 1 and block 2 and assume that coefficients of static and connect friction are equal.

05:35 So coefficients of static and kinetic friction are equal.

05:39 That's the assumption you're making.

05:45 And then the other thing is determine the acceleration, determine the acceleration and this is a of the marble block.

06:07 So that's the acceleration of this marble block.

06:10 Assuming determining determine the acceleration of the marble block assuming a positive x is along the plane with m1 the plane that m1 seats on so what's going to happen is that this is our positive x direction and the block has a weight and this weight happens to be our mg so we want to be careful with the dimensions right here so you know this was this was our angle and this angle right here was alpha so we just want to make sure that give it some space and this other angle was beta and so our weight right here is mg or rather m.

07:21 M1g, our weight is m1g...

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