Auniform 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?

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

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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