Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
A thin uniform wire is bent to form the two equal sides AB and AC of the triangle ABC ,where AB=AC=5cm.The third side BC=6cm, is made from uniform wire oftwice the density of the first but hte thickness of BC is same as that of AB and AC. Find the distance of the centre of mass from A.
The three rods form a triangle as shown below:
weight is concentraed at centre of each rod
let mass of AB=AC=d.5.A ,its C.O.M are (4.5 , 2) and (1.5 , 2)
let mass of BC=2d.6.A ,its C.O.M are (3 , 0)
COMx = (5*4.5 +5*1.5+3*12)/(5+5+12) = 3
COMy = (5*2 +5*2+0*12)/(5+5+12) = 0.91
distance P(3 , 0.91) and A(3 , 4) is 3.09m
--
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.Regards,RameshIIT Kgp - 05 batch
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !