Question

A two-step pulley system is shown in figure. What will be the acceleration $a_p$ of the falling weight P? If P=90 N, Q=135 N and $r_1=2r_2$, Neglect friction and inertia of the two-step pulley.

• A. 3.48 $m/s^2$
• B. 1.78 $m/s^2$
• C. 3.78 $m/s^2$
• D. 3.56 $m/s^2$

Answer 1

The correct option is B 1.78 $m/s^2$

Assume that the acceleration of the weight Q on pulley of radius $r_2$ is $a_q$. Let during any time interval t, the rotation of the pulley is $\theta$. Then we have

$\theta =\frac{l_1}{r_1}=\frac{1_2}{r_2}$

So, $l_1{r}_2=l_2{r}_1$

$l_1\times r_2=l_2\times 2r_2$

$l_1=2l_2$

Now differentiating with respect to time,
$\frac{d{l}_1}{dt}=2\frac{d{l}_2}{dt}$

$\because$ $l_{1}and{l}_2$ are length of ropes Pand Q respectively,
$\frac{d{v}_p}{dt}-2\frac{d{v}_q}{dt}$

$a_p=2a_q$
Writing moment equation of dynamic equilibrium
$(Q+\frac{Q}{g}\times a_q)\times r_2=(P-\frac{P}{g}\times a_p)\times r_1$

$(135+\frac{135}{9.81}\times \frac{a_p}{2})\times r_2$=$(90-\frac{90}{9.81}\times a_p)\times 2r_2$

$135+\frac{135}{9.81}\times \frac{a_p}{2}=180-\frac{180}{9.81}\times a_p$

$\left[\frac{\left(\frac{135}{2}\right)+180}{9.81}\right]a_p=180-135$

$a_p=\frac{45\times 9.81}{247.5}=1.7836m/s^2$