Problem 2.3 on the left side of the balance shown could o00000000 00000oooooo0oo

Question

Problem 2.3 on the left side of the balance shown could o00000000 00000oooooo0ooo000o ooO000000000000000O you hang two square nuts from the same hole so that both sides will balance? 2 B. Suppose that one of the square nuts on the left lease side were moved one hole to the right. What must be done with the other square nut on the left side of the beam so that the beam remains balanced? S What if the square nut that was moved to the right were moved another hole to the right? C. On the basis of your answers to part B predict whether the sides of the beams below will balance. In eac h case, show the steps by which you arrived at your answer. 000oooooooooo0000OO oooooooooooooooo00o O00oOOoooooooooooOO 0o000oooo0000OoooOo O000oOOOoogoooopoooooOo00O0000 Oo00oo00OO0000ooo0o Copyright John Wiley &Sons, Inc. McDermott &P.E.G., U.Wash./Physics by Inquiry

Explanation / Answer

Let mass of each square nut = m

and let distance between two consucetive holes be x.

let the middle hole of the balance is at origin.  ( count central/middle hole as zero)

then on right side, the position of square nut = 8x (as it is situated on 8th hole of right side. we count middle hole as zero.)

we have to add two square nuts (mass = 2m) on left side.

i,e, we have to add of mass of 2m on left side, so that it will balance.

form center of mass, we know that

mass times distance will be constant for both sides.

M1X1 = M2X2 -----------------(1)

Here M1 = 2m (left side)

M2 = m (right side)

X2 =8x

hence X1 = M2X2/M1 = 8x*m/2m = 4x

Hence the two square nuts should be hang on 4th hole from left side of center. (excluding center hole)

2.As per question, if one square nut is moved to one hole right, then new position of one square note of left side is 3x.

let another nut will be kept at distance y from the center to keep the balance.

we have to calculate y.

from eqn(1)

m*3x+m*y = m*8x

my = 8mx-3mx= 5mx

y = 5x

hence another square nut should be moved one hole left on the left side.

hence positions of two nuts on left side will be on 3rd and 5th hole from the center. ( count central/middle hole as zero).

c.

for fig.a, on left side, nuts are 7th hole and 1st hole.

m*7x+m*x = 8mx

on right side two nuts are on 4th hole

2m*4x =8x

hence it will balance.

for fig.b, on left side,

two nuts are on 3rd, one nut on 5th and another nut on 7th

hence 2m*3x+m*5x+m*7x = 18mx

on right side,

mx+4mx+5mx+8mx = 18mx

hence it will balance.

For fig C=

for left side

8mx+3mx+2mx+x = 14mx

for right side,

2mx+5mx+8mx = 15mx

hence, it will not balance.

for fig.d,

for left side,

8mx+3mx+4mx = 15mx

for right side

3mx+12mx = 15mx

hence it will balance.

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