FROM THE FBD WE CAN GET THAT 2TSIN60=4g=40..SOLVING THIS T=40/√3
A rod OA of mass 4 kg is held in horizontal position by a massless string AB as shown in the figure,Length of the rod is 2 m .Find :
a.tension in the string
b.net force exerted gy the hingre on the rod
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4 Answers
Abhishek Priyam
·2009-07-06 10:53:51
balancing torque about hinge.
2*Tsin(60)=mg
T=40/√3
for (b)
(conventional way..)
let req force be Fx and Fy
Fx=Tcos(60)=20/√3
Fy+Tsin(60)=mg
Fy=20
F2=400+400/3
F2=1600/3
F=40/√3
Physics Wallah
·2019-02-01 08:48:26
Philip Calvert
·2009-07-05 01:09:32
Tsin60 = mg/2
T √3 = 40
this should give u (a)
b) F = √(Tcos60)2 + (mg/2)2