A block of mass m1=2.00 kg and a block of mass m2= 6.00 kg...

A block of mass $m_{1}=2.00 \mathrm{~kg}$ and a block of mass $m_{2}=$ $6.00 \mathrm{~kg}$ are connected by a massless string over a pulley in the shape of a disk having radius $R=0.250 \mathrm{~m}$ and mass $M=10.0 \mathrm{~kg}$. These blocks are allowed to move on a fixed block-wedge of angle $\theta=30.0^{\circ}$, as shown in Figure $\mathrm{P} 10.39 .$ The coefficient of kinetic friction for both blocks is $0.360 .$ Draw free-body diagrams of both blocks and of the pulley. Determine (a) the acceleration of the two blocks and (b) the tensions in the string on both sides of the pulley.

Key Concepts

Tension and Constraint Relationships

The tension in the string is not necessarily the same on both sides of the pulley when the pulley has mass and rotational inertia. The massless, inextensible string imposes a constraint linking the accelerations of the blocks, while the pulley’s dynamics require a torque balance that relates the difference in tensions to the pulley's angular acceleration. This concept is essential for correctly formulating and solving the system of equations governing the motion.

Rotational Dynamics

Rotational dynamics deals with the rotation of bodies and how torques affect their angular acceleration. For the pulley modeled as a disk, its rotational inertia must be considered. The moment of inertia for a disk (I = ½MR²) and the relationship between angular acceleration and linear acceleration (a = ?R) are crucial when analyzing how the tensions on either side of the pulley contribute to its rotational acceleration.

Free-Body Diagrams

Free-body diagrams are graphical representations that isolate a body to depict all the forces acting on it. In this context, drawing free-body diagrams for the blocks and the pulley helps identify forces such as gravitational forces, normal forces, frictional forces, and tensions in the string. These diagrams are essential for correctly applying Newton's Laws.

Newton's Second Law

Newton's Second Law is fundamental to analyzing the motion of objects. In both linear and rotational forms, it relates the net force (or net torque) acting on an object to its acceleration (or angular acceleration) multiplied by its mass (or moment of inertia). This principle is used to set up the equations of motion for each block and the pulley, allowing one to solve for unknown accelerations and tensions.

Kinetic Friction

Kinetic friction is the resistive force that opposes the relative motion of two sliding surfaces. In problems involving moving blocks, the kinetic friction force is calculated as the product of the coefficient of kinetic friction and the normal force. This frictional force plays a key role in determining the net force along the direction of motion for each block.

Transcript

00:01 For this problem on the topic of rotation, we have two blocks of mass 2 and 6kg respectively, connected by a massless string over a pulley in the shape of a disc.

00:11 The pulley has a radius of 0 .25 meters and a mass of 10kj.

00:16 Now, the blocks are allowed to move on a fixed block wedge of angle 30 degrees, as we can see in the figure.

00:25 The coefficient of kinetic friction for both blocks is 0 .36.

00:29 We first want to draw free body diagrams of the blocks and pulley, and then determine the acceleration of the two blocks, as well as the tensions in the string on both sides of the pulley.

00:40 Now we've drawn the free body diagrams for mass 1, the pulley, and mass 2.

00:47 Now from these, we look at m1 first, and the sum of the vertical forces must equal m -a -y by newton's second law.

01:03 And so on m1, we have the normal force, n1 minus the weight of mass 1, which is m1g, is equal to 0.

01:20 And so, n1 is equal to m1g, which is 19 .6 neutrons.

01:34 And the frictional force acting on block 1, fk1, is equal to the.

01:40 Coefficient of kinetic friction muke x times n1 which is 7 .06 newtons and if we take the sum of the horizontal forces or the forces along the x direction again this must equal to m a x from newton's second law and we get minus 7 .06 newtons plus the tension t1 is equal to the mass 2 kg is times the acceleration a.

02:20 So the acceleration is purely horizontal for block 1.

02:25 And we'll call this equation 1.

02:27 There's two unknowns here.

02:32 Now for the pulley, we have the sum of talks must equal to i alpha, the angular form of deuton's second law.

02:45 And so minus t1 times r, plus t2 times r must equal a half m r squared, which is the moment of inertia i, times the angle acceleration is acceleration a divided by r.

03:07 So from here we get minus t1 plus t2 is equal to a half times 5 kg times acceleration a.

03:24 And so we'll call this equation 2...

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