Chapter 1, Problem 1.1P

To determine

(a)

To calculate:

The block's position as a function of time for Oxy axes and the time taken by the block to reach the bottom.

Answer to Problem 1.1P

Position of the block in terms of (x,y) = $\left(\left(\frac{g\sin \theta -\mu g\cos \theta }{2}\right)t^2,0\right)$

Time taken by the block to reach the ground $t=\sqrt{\frac{2l}{g\sin \theta -\mu g\cos \theta }}$

Explanation of Solution

Given:

A block of mass msliding down on an inclined plane making angle θ with respect to the horizontal surface.

Formula used:

Using equation of motion,

  $s=ut+\frac{1}{2}a{t}^2$

s = distance travelled by the block

u = initial velocity

t = time taken by the block

a = acceleration of the block

Calculation:

 

 

Forces on the block along x-axis,

Since the block is moving in x direction therefore, by the equation of motion.

  $mg\sin \theta -f_x=ma\mathrm{.......}(1)$

  $mgsin\theta$ is the component of mg along x-axis

  $f_x$ is the frictional force

m is the mass of the block

a is the acceleration of the block

by Newton's law of motion,

  $N=mg\cos \theta$

N = normal force

  $mgcos\theta$ = vertical component of mg along y-axis

force of friction is given by

  $f_x=\mu N\mathrm{.......}(2)$

Put the value of N in equation (2)

  $f_x=\mu mg\cos \theta$

Put the value of f_xin equation (1)

  $\begin{array}{l}mg\sin \theta -\mu mg\cos \theta =ma\\ g\sin \theta -\mu g\cos \theta =a\mathrm{......}(3)\end{array}$

By the equation of motion

  $x=u_{x}t+\frac{1}{2}a{t}^2$

  $u_x\text{ = 0}$ = initial velocity of the block

  $x=\frac{1}{2}a{t}^2$

Put the value of a from equation (3)

  $\frac{1}{2}\left(g\sin \theta -\mu g\cos \theta \right)t^2\mathrm{.......}(4)$

Since there is no motion in vertical direction,

Therefore, y =0

Time taken by the block to reach the ground.

Here, distance covered by the block is l.

Therefore, put $x=l,$ in equation (4)

  $\begin{array}{l}l=\frac{1}{2}\left(g\sin \theta -\mu g\cos \theta \right)t^2\\ t=\sqrt{\frac{2l}{\left(g\sin \theta -\mu g\cos \theta \right)}}\end{array}$

Conclusion:

Therefore, position of the block can be defined in $\left(x,y\right)$ as $\left(\left(\frac{g\sin \theta -\mu g\cos \theta }{2}\right)t^2,0\right)$

Time taken by the block to travel distance l, $t=\sqrt{\frac{2l}{\left(g\sin \theta -\mu g\cos \theta \right)}}$

To determine

(b)

To calculate:

The block's position as a function of time for Ox'y' axes and the time taken by the block to reach the bottom.

Answer to Problem 1.1P

Position of the block in terms of $\left(x’,y’\right)=\left(\frac{1}{2}a\cos \theta t^2,-\frac{1}{2}a\sin \theta t^2\right)$

Time taken by the block to reach the ground, $t=\sqrt{\frac{2l}{g\sin \theta -\mu g\cos \theta }}$

Explanation of Solution

Given:

A block of mass m sliding down on an inclined plane making angle θ with respect to the horizontal surface.

Formula used:

Using equation of motion,

  $s=ut+\frac{1}{2}a{t}^2$

s = distance travelled by the block

u = initial velocity

t = time taken by the block

a = acceleration of the block

Calculation:

 

Distance travelled in x-direction

  $\begin{array}{l}x=l\cos \theta =u_{x}t+\frac{1}{2}{a}_x{t}^2\\ \text{where, }{a}_x\text{is the component of acceleration in x direction}\\ a_x=a\cos \theta \\ u_x=\text{ initial velocity = 0}\\ x=l\cos \theta =\frac{1}{2}a\cos \theta t^2\mathrm{.......}(1)\end{array}$

Distance travelled in y direction

  $\begin{array}{l}y=-l\sin \theta =u_{y}t+\frac{1}{2}{a}_y{t}^2\\ \text{where, }{a}_x\text{is the component of acceleration in x direction}\\ a_y=-a\sin \theta \\ u_y=\text{ initial velocity = 0}\\ y=-l\sin \theta =-\frac{1}{2}a\sin \theta t^2\mathrm{.......}(2)\end{array}$

Position of the block in $\left(x’,y’\right)=\left(\frac{1}{2}a\cos \theta t^2,-\frac{1}{2}a\sin \theta t^2\right)$

Time taken by the block to reach the ground

Adding equation (1) and (2)

  $\begin{array}{l}l\cos \theta -l\sin \theta =\frac{1}{2}a\cos \theta t^2-\frac{1}{2}a\sin \theta t^2\\ 2l(\cos \theta -\sin \theta )=a{t}^2(\cos \theta -\sin \theta )\\ 2l=a{t}^2\\ t=\sqrt{\frac{2l}{a}}\\ \text{put }a=g\sin \theta -\mu g\cos \theta \text{ from equation (3) of part (a)}\\ t=\sqrt{\frac{2l}{g\sin \theta -\mu g\cos \theta }}\end{array}$

It is clear from the solution of part (a) and (b), that time taken by the box is same to travel the distance l.

Taking $Ox’y’$ axes is less convenient because it is difficult to draw free body diagram for the object and so the forces and their directions.

Conclusion:

For $Ox’y’$ axes, time taken by the block to reach the ground $t=\sqrt{\frac{2l}{g\sin \theta -\mu g\cos \theta }}$

For $Ox’y’$ axes, position of the block in $\left(x’,y’\right)$ is given as $\left(\frac{1}{2}a\cos \theta t^2,-\frac{1}{2}a\sin \theta t^2\right)$