Urgent! I don\'t understand how to do this. You will be allowed three submission

Question

Urgent! I don't understand how to do this.

You will be allowed three submissions for this question.

1. In procedure 3: suppose the ruler is asymmetrical, balancing at the 66.7 cm mark. The ruler is now supported at the 44.5 cm mark, and a mass of 310 g is placed at the 16.7 cm mark. If the system is now in equilibrium, find the mass of the ruler.
mruler = _____ g

2. In procedure 4: suppose a symmetrical meter stick of mass 245 g is free to rotate about a fulcrum at the 45.3 cm mark. A mass of 480 g is hanging at the 56.8 cm mark. Mass M is placed at the 26.0 cm mark, and the ruler balances. Find M.
M =  _____ g

Explanation / Answer

here,

centre of mass is at 66.7 cm

let the mass of the ruler be m

when the 0.310 kg mass is laced at 16.7 cm,

the centre of mass is at 44.5 cm

therefore,

the centre of mass X is

X(m1 + m2) = x1 * m1 + x2 * m2

m*66.7 + 0.310 * 16.7 = 44.5 (m+0.31)

m = 0.388 kg

the mass of the ruler is 0.39 kg

please ask seprate question in seprate posts

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