1133m2 = g/3 down
m1 = g/3 up and 4g/15 right
M = 4g/15 right
T = 4g/3
Himanshu Giria How can the unit of tension b that if g
- Himanshu Giria of g
7 Answers
486acc. of m2 g/3 downward.
acc. of m1 g/3 upward.
acc. of M 2g/3 rightward.
tension in the joining m1&m2 is 4g/3
- Himanshu Giria I think the tension ans is dimensionally wrong .
Upvote·0· Reply ·2013-07-15 17:14:23
- Himanshu Giria Plz show how did u get the third ans
- Anurag Ghosh No dude i think its correct.....
Akash Anand Wrong answer....may be you have done some mistakes..or ignored some force..check it again.
11331161All of you have got the wrong answer. This is a very conceptual question of Constraint Relation. All I can give you is a hint for now. Acceleration of 2 kg block and 1 kg block in vertical direction will not be the same (in magnitude). If we take acceleration of 1 kg block as a1 (in upward direction) and acc. of 2 kg as a2 (downward) and acc. of 4 kg as a3 in rightward then constraint relation will be: a2=a1+a3 (Think why??)
- Shreya Sharma Will the pulleys also exert some normal force?
996a1=107 m/s2 (for block of mass 1kg)
a2=307 m/s2 (for block of mass 2kg)
a3=207 m/s2 (for block of mass 4kg)
T=807N
51Are the F.B.Ds correct.
I'm having difficulty in finding the constraint relation .
51
Are the F.B.Ds correct?
I'm having difficulty in finding the constraint relation .
- Akash Anand FBD for the bigger mass and m2 is not correct.