0

Guest

Courses New

A smooth ring of mass M can slide on a fixed horizontal rod.A massless string tied to the ring passes over a fixed smooth Pulley of mass M and carries a block of mass 2m.At a instant the string between ring and Pulley makes an angle 30° with horizontal.ring acceleration of block and ring,normal force due to ring on the rod,reaction force on the Pulley due to support

A smooth ring of mass M can slide on a fixed horizontal rod.A massless string tied to the ring passes over a fixed smooth Pulley of mass M and carries a block of mass 2m.At a instant the string between ring and Pulley makes an angle 30° with horizontal.ring acceleration of block and ring,normal force due to ring on the rod,reaction force on the Pulley due to support

Grade:12th pass

1 Answers

Ankit Raj
39 Points
6 years ago
I must say that image will explain more clearly but no problem. For 2M mass 2Mg-T=2Ma ,for ring Tcos30°=Ma` one equation from constraint relationship that length of string remains same so acceleration of both ends will be same therefore a`cos30°=a Answers would be accel.of 2M (3/5)g accel. Of ring (2√3/5)g Tension (4/5)Mg. Net supporting force on pulley √3T as there will be a horizontal as well.Normal reaxn on ring (2/5)Mg

Think You Can Provide A Better Answer ?

Other Related Questions on Mechanics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More