- #1
dyn
- 773
- 62
- Homework Statement
- Massless pulley . massless string with block of mass m1 at one end . block of mass m2 at other end. If pebble of mass m is placed on block of mass m2 , what is the force exerted by the pebble om block of mass 2 ?
- Relevant Equations
- Newton's 2nd law ; F=ma
Assume m2>m1 and take positive direction as downwards. String is inextensible so acceleration of block up = acceleration of block +pebble down =a
I used the following 2 equations to obtain a ;
-m1a = m1g - T , (m2+m)a = (m+m2)g - T
which gives a = g(m+m2-m1)/(m+m1+m2)
I then constructed a free-body diagram for the pebble which has its weight acting down and the normal reaction from the block acting upward.
Using ma= mg - N ; i arrived at N = 2m1mg/(m+m1+m2
I then said that the force exerted by pebble on m2 is equal and opposite to the normal reaction , N
Where am i going wrong ? Thanks
I used the following 2 equations to obtain a ;
-m1a = m1g - T , (m2+m)a = (m+m2)g - T
which gives a = g(m+m2-m1)/(m+m1+m2)
I then constructed a free-body diagram for the pebble which has its weight acting down and the normal reaction from the block acting upward.
Using ma= mg - N ; i arrived at N = 2m1mg/(m+m1+m2
I then said that the force exerted by pebble on m2 is equal and opposite to the normal reaction , N
Where am i going wrong ? Thanks
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