we have that from the Question, it can be said that distance x from the left-hand end of the bar is
x=1.834m
From the Question we are told
A uniform metal bar is 5.00 m long and has mass 0.300 kg. The bar is pivoted on a narrow support that is 2.00 m from the left-hand end of the bar. What distance x from the left-hand end of the bar should an object with mass 0.900 kg be suspended so the bar is balanced in a horizontal position?
Generally the equation for Balanced torque is mathematically given as
T=0.3/5*3
T=0.18kg
Therefore
The clockwise torque is
T_c=0.18*1.5g
And
The anti-clockwise torque is
T=0.3/5*2g
T=0.12
Hence
0.18*1.5g=0.3/5*2g+0.9kxg
Therefore
x=1.834m
Therefore
distance x from the left-hand end of the bar is
x=1.834m
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The position of the 0.9 kg mass to balance the metal bar is 1.83 m.
The given parameters;
The center of gravity of the metal bar = 2.5 m
A sketch of weight on the metal bar;
|-----------2 m-----------|--- 0.5 m-----|
0---------------------------Δ--------------------------------------------------
P ↓|-------- x----------| ↓
0.9 kg 0.3 kg
Take moment about the pivot point;
0.9x = 0.3(0.5)
0.9x = 0.15
[tex]x = \frac{0.15}{0.9} \\\\x = 0.167 \ m[/tex]
2 m - P = 0.167 m
P = 2m - 0.167 m
P = 1.83 m
Thus, the position of the 0.9 kg mass to balance the metal bar is 1.83 m.
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