An object with a mass of 24 kg is hanging from an axle with a radius of 16 m. If the wheel attached to the axle has a radius of 48 m, how much force must be applied to the wheel to keep the object from falling?

2 Answers
Jan 8, 2018

The force is =78.4N

Explanation:

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The load is L=24gN

The radius of the axle is r=16m

The radius of the wheel is R=48m

The effort is =FN

The acceleration due to gravity is g=9.8ms^-2

Taking moments about the center of the axle

F*48=24g*16

F=(24g*16)/48=78.4N

The force is F=78.4N

Jan 8, 2018

78.4052N to 4 decimal places

Explanation:

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Did you know that units of measurement can be manipulated in the same way that numbers can. If you are ever not sure what to do with the numbers look at what you have to do to the units to give you your target.

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The mass of 24 kg converted to downward force in Newtons is:

24cancel(kg)xx9.80065 N/(cancel(kg)) = 235.2156N to 4 decimal places

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color(blue)("Answering the question")

Presenting opposing forces about the shaft centre line of the shaft as if it is a fulcrum we have:

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If this condition is not 'in equilibrium' (all balanced out) then the system would be in motion.

Taking moments about the fulcrum

48x_1Kgm=16xx24 Kgm

But force is measure in Newtons by changing the units of measurement we have:

48x_2Nm=16xx235.2156 Nm

Divide both sides by 48m (gets x Newtons on its own)

(48m)/(48m)xx x_2N=(16xx235.2156)/48 color(white)("d")(Ncancel(m))/cancel(m)

But (48m)/(48m) = 1 and 1xx xN=xN

x_2N=78.4052N to 4 decimal places