- #1
Kudo Shinichi
- 109
- 1
HELP!Statics problem
A thin rod AB of length L=0.5m and mass m=1.0kg makes an angle of theta=60 degree with the horizontal. It is held in position by a horizontal light string attached at A as shown. A mass M is suspended from A by a second string. If the coefficient of friction between the rod and the ground at the point of contact B is mu(s)=0.50, what is the largest mass M that can be suspended such that the rod does not slip.
Diagram:
http://tinypic.com/view.php?pic=121slk5&s=4
we need to use the equilibrium conditions to solve the problem. therefore, Sigma(Fx)=0,Sigma(Fy)=0 and Sigma(torque)=0
Sigma(Fy)=FN-mg=0
therefore, normal force=mg=1*9.8=9.8N
Sigma(Fx)=frictional force-exerted force=0
this is what I know for solving the problem so far...I know that I have to somehow relate the rod with the hanging mass.
Ignoring the ways I attempted above, I think that i can get the mass by
M=1.0kg*cos(30o)
but I think it is not the correct answer.
Can anyone help me with it? thank you very much.
Homework Statement
A thin rod AB of length L=0.5m and mass m=1.0kg makes an angle of theta=60 degree with the horizontal. It is held in position by a horizontal light string attached at A as shown. A mass M is suspended from A by a second string. If the coefficient of friction between the rod and the ground at the point of contact B is mu(s)=0.50, what is the largest mass M that can be suspended such that the rod does not slip.
Diagram:
http://tinypic.com/view.php?pic=121slk5&s=4
The Attempt at a Solution
we need to use the equilibrium conditions to solve the problem. therefore, Sigma(Fx)=0,Sigma(Fy)=0 and Sigma(torque)=0
Sigma(Fy)=FN-mg=0
therefore, normal force=mg=1*9.8=9.8N
Sigma(Fx)=frictional force-exerted force=0
this is what I know for solving the problem so far...I know that I have to somehow relate the rod with the hanging mass.
Ignoring the ways I attempted above, I think that i can get the mass by
M=1.0kg*cos(30o)
but I think it is not the correct answer.
Can anyone help me with it? thank you very much.
Last edited: