A block of mass m1=2.00 kg and a block of mass m2=6.00 kg are connected by a massless string ove

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 solid disk having radius $R=0.250 \mathrm{m}$ and mass $M=10.0 \mathrm{kg} .$ The fixed, wedge-shaped ramp makes an angle of $\theta=30.0^{\circ}$ as shown in Figure P10.72. The coefficient of kinetic friction is 0.360 for both blocks. (a) Draw force diagrams of both blocks and of the pulley. Determine (b) the acceleration of the two blocks and (c) the tensions in the string on both sides of the pulley.

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 solid disk having radius $R=0.250 \mathrm{m}$ and mass $M=10.0 \mathrm{kg} .$ The fixed, wedge-shaped ramp makes an angle of $\theta=30.0^{\circ}$ as shown in Figure P10.72. The coefficient of kinetic friction is 0.360 for both blocks. (a) Draw force diagrams of both blocks and of the pulley. Determine
(b) the acceleration of the two blocks and (c) the tensions in the string on both sides of the pulley.

Key Concepts

Tension Variation

In systems with a massive and rotating pulley, the tension in the string can differ on either side of the pulley. This variation is necessary to produce a net torque on the pulley, and it must be accounted for when applying Newton’s laws to resolve the system’s dynamics.

Constraint Relations in Pulley Systems

Constraint relations refer to the mathematical relationships that arise from the inextensibility of the connecting string in a pulley system. They link the translational motion of the masses with the rotational motion of the pulley, ensuring that the linear acceleration of the masses is consistent with the angular acceleration of the pulley.

Moment of Inertia

The moment of inertia is a property of a body that quantifies its resistance to angular acceleration about an axis. For rotational systems like a solid disk pulley, knowledge of its moment of inertia is crucial for relating the applied torques to the angular acceleration.

Kinetic Friction

Kinetic friction is the resistive force that opposes the motion of objects sliding relative to each other. It is proportional to the normal force and is characterized by a coefficient that depends on the interacting surfaces, critically affecting the dynamics of the blocks.

Newton’s Second Law

Newton's Second Law states that the net force acting on an object is equal to its mass times its acceleration. It is applied in both translational and rotational dynamics to determine how forces influence the linear and angular accelerations of objects.

Free-Body Diagrams

Free-body diagrams involve drawing simplified representations of objects where all the forces acting on them are shown. This visualization helps in setting up the equations of motion by clearly identifying all acting forces such as gravity, normal forces, friction, tension, and applied forces.

Rotational Dynamics

Rotational dynamics concerns the study of torques and angular accelerations in rotating bodies. It connects the sum of torques with the product of moment of inertia and angular acceleration, making it essential when dealing with pulleys or any rotating system.

Transcript

00:01 See the figure force diagram at right for m1 summation of f y is equal to m of a y gives us plus n minus m g is equal to zero hence n1 is equal to m g for f of k1 which is equal to muo k n one now submission of f x is equal to m a x for f of k one which is equal to muo k n one now submission of f x is equal to m a x for along x direction is equal to mass into acceleration along x direction gives minus f of k1 plus t1 is equal to m1 a which is equation for the pulley submission summation is equal to i alpha minus t 1 r plus t 2r is equal to 1 by 2 m r square multiply with a upon capital r now minus t plus t 2 minus t plus t 2 t 1 plus t 2 is equal to 1 by 2 m r multiply with a upon capital r which tends to minus t 1 plus t 2 is equal to half of m 2 form m 2 now for m 2 is equal to m 2 we have plus n 2 minus m 2 g course of theta is equal to 0 hence n2 is equals to m2 g coss of theta now f of k2 is equal to mu k into n2 from here similarly minus of f of k 2 minus of f of k 2 is equals to mu k n2 mu k n2 now minus of f of k 2 is equal to m m u k n2 now minus of f of k 2 is equal to minus of f of k 2 of k2 minus t 2 plus m2 g sine of theta is equals to m2a now in part b add equations this is equation number 3 here this is equation number 2 t is equals to half m a and this is equation number 1 here this is equation number 1 now what may be note to do in part b we have to add 1, 2 and 3 and substitute the expression for f k1 and n1 and minus f k 2 and n2.

02:49 Now minus of f of k 1 plus t 1 plus minus of k 1 plus t 2 minus of f of k 2 minus f of k 2 minus 2 plus m 2 g sine of theta is equal to n1 a plus half of m a plus m a plus m 2 plus m2 a minus f of k1 minus f of k2 plus m2 plus m2 g sign of theta we get after cutting them is equal to m1 plus m2 plus half 1 by 2 of capital m and a is common here common now we get minus of mu k m1 g minus of new k m1 g minus of new k m2 g o's of theta plus m2g sine of theta is equal to m1 plus m2 plus half of capital m into a now from here we can find the value of acceleration which is m2 into sin theta minus mu k oph of theta minus mu of k m1 divided by m1 plus m2 plus m2 plus 1 by 2 of capital m multiply with g.

04:16 After putting the values of a, we have a is equals to m2 is 6kg, sine theta, sine 30 degrees, minus nu kkkkk, which is 0 .360 cost 30 degree minus mu k, which is 0 .360, multiply with m1, which is 2kg.

04:42 Whole is multiplied with g and this is divided by a which is equal to divided by m1 which is 2 kg plus m2 which is 6 kg plus m1 plus 6kg plus m1 plus 6 kg plus half of 10 .0 kilogram we get the acceleration is equals to 0 .309 meter per second square now in part c from equation 1, we get minus of fk1 plus t1 is equal to m1a tends to t1, which is equal to 2 kilogram multiplied 0 .309 meter per second square plus 7 .06 newton...

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