Question

Write clearly, explain well.

Problem 1: Masses, friction inclines, and a pulley (5 points) A solid cylinder of mass (m) and radius (r) is on a plane inclined at an angle (0). The cylinder has a string wrapped around it and attached to a wall at the top of the incline. The string runs parallel to the incline. There is a coefficient of kinetic friction (uk) between the cylinder and the plane. The cylinder begins to slide and unravel the string. What is Tension Force of the string? Knowns: 0, m, R, Hk, and g (acceleration due to gravity) No numbers, give all answer in terms of the knowns

Answer 1

on mgsine mgcos mg

Let T = tension

fs = friction force

by using formula

mgsin$\Theta$ -fs - T = ma .........................eq (i) ... (where a = acceleretion & m = mass of sylinder)

torque about center of cylinder $\tau$ = T*r + fs *r ..........(where r = radius of cylinder)

But
i=19
..........(where = moment of inrtia $I = \frac{1}{2}mr^{2}$ & $\alpha$ = angular acceleretion = a/r)

So ,  $I\alpha =$

$\frac{1}{2}mr^{2}*\frac{a}{r}=T*r + fs *r$

$\frac{1}{2}mra=T*r + fs *r$

$\frac{1}{2}ma=T+ fs$

$a=\frac{2(T+ fs)}{m}$

But $f_{s}= \mu _{k}N$ ..........................(where N= normal force = mgcos$\Theta$)

So ,$f_{s}= \mu _{k}mgcos\Theta$

Or ,  $a=\frac{2(T+ \mu _{k}mgcos\Theta )}{m}$

Put value of a in eq (i)

$mgsin\Theta -f_{s}-T =m\frac{2(T+ f_{s} )}{m}$

$mgsin\Theta -f_{s}-T =2(T+ f_{s} )$

$mgsin\Theta -f_{s}-T =2T+ 2f_{s}$

$3T=mgsin\Theta -3f_{s}$

But $f_{s}= \mu _{k}N$ ..........................(where N= normal force = mgcos$\Theta$)

So ,$f_{s}= \mu _{k}mgcos\Theta$

Or ,  $3T=mgsin\Theta -3\mu _{k}mgcos\Theta$

$T=\frac{mgsin\Theta -3\mu _{k}mgcos\Theta}{3}$ ..........................Ans