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

Weights of 1g,2g,.....,100g are suspended from 1cm,2cm,.....,100cm mark respectively. Where i t should be suppoted so, that the the system is in equlibrium??

Weights of 1g,2g,.....,100g are suspended from 1cm,2cm,.....,100cm mark respectively. Where it should be suppoted so, that the the system is in equlibrium??

Grade:

1 Answers

Kaustubh Nayyar
27 Points
6 years ago
the point must be the centre of mass
it can be calculated as
COM =  ( 1+22+3+ …......+1002 ) / ( 1+2+3+4+..........+100 ) 
         =  { (100)(101)(201) / 6 } / { (100)(101)/2}  
           = 201/3  = 67 cm for origin  if all the weights are at x axis
hence at 67cm from the origin or 66 cm from the first particle

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free