A 3.5-kg object placed on an inclined plane (angle 30 above the horizontal) is connected by a st

step 1 Now, Question 77 is super interesting. It can get quite involved if you're not careful. And so I

A 3.5-kg object placed on an inclined plane (angle 30 above...

Key Concepts

String and Pulley Systems

This concept deals with the dynamics of connected objects where one object’s motion directly affects the other through a string and a pulley. The string is usually assumed to be massless and inextensible, meaning that the tension is the same throughout, and the acceleration of the connected objects is common.

Free-Body Diagram

A free-body diagram is an essential tool used in physics to visually represent all the forces acting on an object. It helps in organizing and decomposing the forces involved, such as gravity, normal force, friction, and tension, which is crucial for setting up the equations of motion.

Inclined Plane Analysis

This concept involves resolving forces into components that are parallel and perpendicular to the surface of the inclined plane. The gravitational force is decomposed into a component that causes the object to slide down the plane and a normal force that acts perpendicular to the plane, which is essential for calculating friction.

Newton's Second Law

This fundamental principle of dynamics states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). It forms the basis for analyzing the motion of objects when forces such as gravity and friction are acting upon them.

Friction

Friction is the resistive force that opposes motion between two surfaces in contact. In dynamics problems, both static and kinetic friction coefficients are used to determine whether an object will start moving or continue moving, with the frictional force typically calculated as the product of the friction coefficient and the normal force.

Transcript

00:02 Now, question 77 is super interesting.

00:04 It can get quite involved if you're not careful.

00:07 And so i'm going to show you a sort of a nice way of thinking about this one.

00:11 But essentially it's another one in the series of blocks on slopes questions.

00:15 And it takes up from last time.

00:17 So we have a 3 .5 kilogram object placed on an inclined plane.

00:22 Once again, angle 30 degrees above the horizontal.

00:26 And it's connected by a string going over a pull into a one kilogram object.

00:29 So once again we just have it's kind of set up as a pulley there with an object here and an object here will model them as ever as particles and determine the acceleration of the system if the coefficient of the static friction between object 1 and the surface of the incline plane is 0 .3 and this is also the coefficient of kinetic friction.

00:54 Ah, frictions got involved and that has complicated matters.

01:01 So, rather generally, let's look at what's going on here.

01:05 So we've got an angle of 30 degrees here, and now the normal reaction force is important because we have a motion of friction.

01:14 I'll come back to that in a second.

01:16 Let's add a m2, we always consider the one on the left to be block two, and we have the weight of block one, m1g, which of course can be split into two components.

01:30 Whose angle here is also 30 degrees.

01:35 Here's our problem.

01:37 The block could move up the slope.

01:39 Block one could move up the slope with block two descending, or block one could be descending the slope and block two ascending.

01:49 This coefficient of friction makes all the difference, and at least thankfully it is also equal to the coefficient of static friction and the coefficient of kinetic friction at the same time.

02:02 We need to determine which way this block is going before we can make any sort of judgment on this.

02:09 And because friction is always opposing motion, so if block one is ascending the slope, then its resistive force is going to come downwards.

02:22 But likewise, if it's descending the slope, then its frictional force will be upwards.

02:28 So actually this makes a bit of a difference.

02:32 This means that we've got to solve this question through logic.

02:37 Let's think of one thing.

02:38 Tension is always in opposition to forces that are naturally acting on the object.

02:48 Our weight essentially have some component that want both of our blocks to move separately, but the tension is drawing them back, and it's always pulling.

03:00 And if we write in equation form that our tension is always pulling, we have to say that it's positive.

03:07 And so let's just assume that tension is positive and then sort of factor out with the bit of logic how we would interpret this.

03:16 So there's two cases to consider.

03:18 Case one is that block one descends.

03:31 I've spelled that right.

03:32 I hope i have.

03:34 Apologies if i haven't to those of you who are looking at this.

03:39 Well if block one is descending, simply it has one out and that means that all the forces that are acting it.

03:49 Of course tension is going to act with both blocks.

03:51 So so what we're really saying is that m1g sine theta or sine 30, which is the parallel component of weight to the slope.

04:09 And we'll subtract from that.

04:12 The coefficient of friction, thankfully, it's the same coefficient, m1, cosine 30, also multiplied by g.

04:21 We should stick a g in here as well.

04:23 This has got to be greater than m2.

04:25 To g...

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