The systems shown below are in equilibrium (with m = 5.50 kg and θ = 28.0°). If

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The systems shown below are in equilibrium (with m = 5.50 kg and θ = 28.0°). If the spring scales are calibrated in newtons, what do they read? Ignore the masses of the pulleys and strings and assume the pulleys and the incline are frictionless.


May 31st, 2015

By looking at and analyzing the position of the masses in the diagram, the universal g constant, 9.81 m/s^2, must be divided by 2 and multiplied by the total kg:
(9.81 m/s^2) / 2 = 4.905 m/s^2
4.905 m/s^2(11.00 kg) = 49.1 kg(m/s^2) = 49.1 N 


Since there are 2 masses:
2(5.5 kg) = 11.00 kg

Now to get the answer, multiply the total kg by the universal g constant, 9.81 m/s^2 because the force is acting downwards:
11.00 kg(9.81 m/s^2) = 98.1 kg(m/s^2) = 98.1 N

May 31st, 2015

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May 31st, 2015
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