A = 59.35cm
B = 196.56g
C = 74.65g
Explanation:
We know,
[tex]x = \frac{L}{\frac{W}{F} +1}[/tex]
and L = x+y
1.
Total length, L = 100cm
Weight of Beam, W = 71.8g
Center of mass, x = 49.2cm
Added weight, F = 240g
Position weight placed from fulcrum, y = ?
[tex]L-y = \frac{L}{\frac{W}{F}+1 } \\100 - y = \frac{100}{\frac{71.8}{49.2}+1 } \\100 - y = \frac{100}{1.46+1}\\\\100 - y = \frac{100}{2.46} \\100-y = 40.65\\\\y = 59.35cm[/tex]
Therefore, position weight placed from fulcrum is 59.35cm
2.
Total length, L = 100cm
Center of mass, x = 47.8 cm
Added weight, F = 180g
Position weight placed from fulcrum, y = 12.4cm
Weight of Beam, W = ?
[tex]47.8 = \frac{100}{\frac{W}{180}+1 }\\\47.8 = \frac{100}{\frac{W+180}{180} } \\\\47.8 = \frac{100 X 180}{W+180}\\ \\47.8W + 47.8 X 180 = 18000\\47.8W = 18000 - 8604\\W = \frac{9396}{47.8}\\ W = 196.56g[/tex]
Therefore, weight of the beam is 196.56g
3.
Total length, L = 100cm
Center of mass, x = 50.8 cm
Position weight placed from fulcrum, y = 9.8cm
Weight of Beam, W = 72.3g
Added weight, F = ?
[tex]50.8 = \frac{100}{\frac{72.3}{F}+1 }\\\ 50.8 = \frac{100}{\frac{72.3+F}{F} } \\\\50.8 = \frac{100 X F}{72.3+F}\\ \\50.8 X 72.3 + 50.8 X F = 100F\\\\3672.84 = 100F-50.8F\\3672.84 = 49.2F\\F = 74.65g[/tex]
Therefore, Added weight F is 74.65g
A = 59.35cm
B = 196.56g
C = 74.65g
