1. The problem statement, all variables and given/known data
A 100 kg slender rod is lifted on the left end until the angle between the ground and the right end is 30 degrees. The right end is still in contact with the ground. The coefficient of friction between the rod and the ground is .5 and the length of the rod is 2.1 m.
2. Relevant equations
∑fx=m(ag)x
∑fy=m(ag)y
∑Mg=Ig∂
Ff=.5Fn
3. The attempt at a solution
I summed the forces in the x direction and got -Ff=m(ag)x
In the y direction I got -981+Fn=m(ag)y but i think the y acceleration of the center of gravity may be zero because of the fact that the bar can't translate up or down due to the contact with the ground. It can only rotate or slip horizontally on the ground. Finally i summed the moments about the center of gravity and got that fn(1.05cos(30))+Ff(1.05sin(30))=36.75∂
I got a final answer of 17.26 rad/s^2 counter clockwise, but i don't think it is correct. Any suggestions?
A 100 kg slender rod is lifted on the left end until the angle between the ground and the right end is 30 degrees. The right end is still in contact with the ground. The coefficient of friction between the rod and the ground is .5 and the length of the rod is 2.1 m.
2. Relevant equations
∑fx=m(ag)x
∑fy=m(ag)y
∑Mg=Ig∂
Ff=.5Fn
3. The attempt at a solution
I summed the forces in the x direction and got -Ff=m(ag)x
In the y direction I got -981+Fn=m(ag)y but i think the y acceleration of the center of gravity may be zero because of the fact that the bar can't translate up or down due to the contact with the ground. It can only rotate or slip horizontally on the ground. Finally i summed the moments about the center of gravity and got that fn(1.05cos(30))+Ff(1.05sin(30))=36.75∂
I got a final answer of 17.26 rad/s^2 counter clockwise, but i don't think it is correct. Any suggestions?
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