This question was for extra credit, I found the center of gravity to be (2.6m,2.

Question

This question was for extra credit, I found the center of gravity to be (2.6m,2.4m). Part b) has zero angular acceleration and t net would equal zero. Part c) I was having trouble on. I might be wrong in all of the above but I tried first at least, I might not have used the right units or something.

I tried to upload a JPG document but it wouldn't let me, so I will try to explain the best I can.

A traingle with three particles m1, m2, and m3. point of origin (0,0) is m3. From m1 to m2 is 5 meters. From m1 to m3 is 5 meters. From m3 to m2 is 6 meters. m2=2m1 and m3=3m1

a) Where is the center of gravity relative to m1?

b) What is the net gravitational torque about this point?

c)What is the ratio of the net gravitational torque to the weight of mass 1, if the above arrangement of masses is pivoted about an axis running through m1 and perpendicular to the arrangement?

Explanation / Answer

A traingle with three particles m1, m2, and m3. point of origin (0,0) is m3. From m1 to m2 is 5 meters. From m1 to m3 is 5 meters. From m3 to m2 is 6 meters. m2=2m1 and m3=3m1

a) Where is the center of gravity relative to m1?

let m1 ia at ( 5 , 0)

m2 is at ( 3. 6 , 4.79)

X (cordinate of gravity) = (m1*5 + m2*3.6 +m1*0)/(m1+m2+m3)

= (5m1 + 7.2m1) /(6m1)

= 2.03

Y = (m1*0 +m2*4.7+m3*0)/(6m1)

= +1.56

centre of gravity relative to m1 = (2.03-5) , (1.56 -0)

= -2.97 , 1.56

b) Torque = m1*g* [ (5-2.03)^2 + (1.56)^2]^0.5 + m2g[ (3.6-2.03)^2 + (4.79-1.56)^2]^0.5

+m3g[1.56^2 + 2.03^2]^0.5

= 3.35m1g + 7.18*m1g + 7.68*m1g

=18.84*m1*g

= 188.4 M1

c) net torque about m1 = m3*g*5 + m2*g*5 =25*m1*g

ratio = 25

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