Two blocks (m1 = 3kg, m2 = 2kg) are connected over a pulley...

Two blocks (m1 = 3kg, m2 = 2kg) are connected over a pulley as shown below. The pulley (C=1/2) has mass M = 2kg and radius R = 60cm. The horizontal surface is rough with coefficient of kinetic friction of 0.7. a) [20 points] Draw a force diagram of each block and the pulley. What is the linear acceleration of blocks? b) [10 points] What is the tension in the rope on either side of the pulley? c) [10 points] Find the magnitude and direction of angular acceleration of the pulley.

Transcript

00:01 We are given here two blocks so in this question we are given here two blocks of the mass m1 which is equal we can weight 3 kilogram and here m2 which is equal to we can weight 2 kilogram i connected over a pulley that connected over a pulley and here suppose pooley that is c which is equal 1 half and the capital m we are given here 2 kilogram capital m is 2 kilograms the radius capital r which is equal we can write 60 cm and the horizontal surface of the coefficient of kinetic friction so we can write coefficient of kinetic friction nuclear which is equal we can write 0 .7 in the first of part of this question here is in the first part of this question we want to find draw the force diagram of each block and the pulley what is a linear as a solution of the block so here we want to find linear acceleration of block so we want to find linear oscillation of the block at a that we want to find let's draw that for the fourth diagram first and finding the answer the first understand that suppose we have a block and the honor block there's a the perpendicular force is acting here we can write m212b of g the downward force is acting here on that direction there is a fk kinetic here here here kind of friction that is the force of friction we are given here in our left direction as a tension t2 and here for a hanging hanging and here suppose this is our hanging block here this is our table block here we can write hanging block hanging block we can write the t1 this is a value is t1 divided by we can write m1 1 1 litre by g the downward force acting is m a and for pulley here we can write for fully so we have a soccer and does it now here capital r divided by we can write value of t1 here we can write value of t2 and here we have alpha now for the table group we can write t2 minus f k which is equal to t2 minus f k which is equal we can write m2 multiple by so this is the value here and now here we can write the t2 minus we can write here first we can write t2 which is into here t2 which is equal to we can write t2 which is equal to m2 multiple by alpha thus we can write mu k m2 by g…

Question

Please give Ace some feedback