A simple weighing machine is made of a uniform bar 125cm long and mass 5kg is pivoted 2.5cm from one end. Find the mass that must be suspended at the end of the long arm so as to balance a mass of 320kg suspended at the end of the short arm.
Answers
Answer:
Considering the weight of the rod to be equally distributed all along the length(that a uniformly distributed load of .6 N/m), we can equate the moments (torque balance) of all weights(along with the said weight of 0.5 N at a distance say x from the fulcrum) about the fulcrum at .45m from one end. This gives the answer as 6 cm.
Another, more simpler, way to go about it is to consider the weight of the rod concentrated at its center and then balance weights. this is also more intuitive. The problem is basically the working of a lever. Move a weight, say x, some distance away from the fulcrum now how much do you need to place another weight y to counter act x.
However, in this case the options are so absurdly simple that the answer can not be anything but B(0.06 m) as a deflection of 5 cm from center of gravity can be easily balance by nearly the same mass at nearly the same deflection.
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While I agree that this is quite simple, the previous answer doesn't actually explain how to arrive at the solution.
First, I think you can see by inspection that the answer for this problem (I am assuming the bar is of uniform construction) is the same as a weightless bar with a weight of .06N placed 50 cm from the fulcrum. That is because the 45cm on the short side exactly balances out the closest 45cm on the long side, and the average distance of the remaining 10cm is 50cm from the fulcrum. The .06N comes from the fact that the length of this "extra" is 10% of the total, so 10% of 0.6N
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A uniform meterstick, supported at the 30.0 cm mark, is balanced when a 0.50 N weight is hung at the 0.0 cm mark so what is the weight of meterstick?
A uniform rod 30 inches long is pivoted at its centre. A 40 pound weight is hung 5 inches from the left end, where must a 50 pound weight be hung to maintain equilibrium?
A meter stick balances horizontally on a knife edge at the 50.0 cm mark. With two 5.00g coins stacked over the 12.0 cm mark, the stick is found to balance at the 45.5 cm mark. What is the mass of the meter stick?
Why is the equilibrium point of a meter rule not a 50 cm mark?
A man who weighs 50.0 N stands 3.00 m from the right end of a board that is supported by two vertical cords attached to either end of the board. The board is 5.00 m long and weighs 10.0 N. What is the tension in each of the two cords?
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Explanation:
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