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Two boxes on opposite ends of a massless board that is 3.0 m long. The board is supported in the middle by a fulcrum. The box on the left has a mass (m1) of 25 kg, and the box on the right has a mass (m2) of 15kg. How far should the fulcrum be positioned from the left side of the board in order to balance the masses horizontally?

a. 0.38 m
b. 1.1 m
c. 0.60 m
d. 1.9 m

Sagot :

hamzaahmeds

Answer:

b. 1.1 m

Explanation:

It is given that the total distance between the masses is equal to the length of the board, which is 3 m. Therefore,

[tex]s_{1} + s_{2} = 3\ m\\\\s_{2} = 3\ m - s_{1}\ --------- eqn(1)[/tex]

where,

s₁ = distance of fulcrum from left mass

s₂ = distance of fulcrum from right mass

In order to achieve balance, the torque due to both masses must be equal:

[tex]T_{1} = T_{2}\\m_{1}s_{1} = m_{2}s_{2}\\(25\ kg)(s_{1}) = (15\ kg)(s_{2})\\\\\frac{15\ kg}{25\ kg}(s_{2}) = s_{1}\\\\using\ eqn(1):\\(0.6)(3\ m - s_{1}) = s_{1}\\1.8\ m = 1.6\ s_{1}\\s_{1} = \frac{1.8\ m}{1.6}[/tex]

s₁ = 1.1 m

Hence, the correct option is:

b. 1.1 m

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