badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12th pass

                        

Three masses m1,m2,m3 are connected with string and placed on smooth table. Now m1 is pulled with force F and common acceleration produced is a. Find individual force on each mass.

2 years ago

Answers : (1)

Susmita
425 Points
							
a=F/(m1+m2+m3)
For each box weight is balanced by tension.
For m3,
T3=m3a=m3F/(m1+m2+m3). This is the force on m3.
Due to equality between action and reaction T1 will act on m2 in left direction.Also T2 will act in the right.So
T2-T1=m2a
or,T2=m2a+T1.Put the values obtained earlier ti get T2.This is the force on m2.
In case of m1 also T2 acts in left and T3 acts in right.So
T3-T2=m1a
T3=T2+m1a is the force on m3.
If it helps please approve.
 
2 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details