wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A two-step pulley system is shown in figure. What will be the acceleration ap of the falling weight P? If P=90 N, Q=135 N and r1=2r2, Neglect friction and inertia of the two-step pulley.


A
3.48 m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1.78 m/s2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
3.78 m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3.56 m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 1.78 m/s2
Assume that the acceleration of the weight Q on pulley of radius r2 is aq. Let during any time interval t, the rotation of the pulley is θ. Then we have

θ=l1r1=12r2

So, l1r2=l2r1

l1×r2=l2×2r2

l1=2l2

Now differentiating with respect to time,
dl1dt=2dl2dt

l1 and l2 are length of ropes Pand Q respectively,
dvpdt2dvqdt

ap=2aq
Writing moment equation of dynamic equilibrium
(Q+Qg×aq)×r2=(PPg×ap)×r1

(135+1359.81×ap2)×r2=(90909.81×ap)×2r2

135+1359.81×ap2=1801809.81×ap

⎢ ⎢ ⎢ ⎢(1352)+1809.81⎥ ⎥ ⎥ ⎥ap=180135

ap=45×9.81247.5=1.7836 m/s2

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Second Law of Motion
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon