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

A mass m is at a distance a from one end of a uniform rod of length l and mass M. The gravitational force on the mass due to the rod is ------------

A mass m is at a distance a from one end of a uniform rod of length l and mass M. The gravitational force on the mass due to the rod is ------------

Grade:12

2 Answers

vikas askiitian expert
509 Points
10 years ago

force on mass m due to rod is equal to force on rod due to mass so   (action equals to reaction)

 i am calculating force on rod due to mass...

let rod is placed in x axis  such that its one end is at a distance x from origin ...mass m is placed at origin...

consider a element dx on rod at a distance x from origin...

mass of this element = dm = (M/L)dx

force on this element due to rod is dF = Gmdm/x2

           dF = Gm(M/L)dx/x2

 integrating both sides

       [F] lim0 to F = -GMm/L (1/x)          lim x to x+L

      F = GMm/x(x+L) 

approve if u like it

Rishi Sharma
askIITians Faculty 646 Points
one year ago
Dear Student,
Please find below the solution to your problem.

force on mass m due to rod is equal to force on rod due to mass so (action equals to reaction)
i am calculating force on rod due to mass...
let rod is placed in x axis such that its one end is at a distance x from origin ...mass m is placed at origin...
consider a element dx on rod at a distance x from origin...
mass of this element = dm = (M/L)dx
force on this element due to rod is dF = Gmdm/x2
dF = Gm(M/L)dx/x2
integrating both sides
[F] lim0 to F = -GMm/L (1/x) lim x to x+L
F = GMm/x(x+L)

Thanks and Regards

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