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CaHeK987 [17]
3 years ago
15

Person is lifting a 250 N dumbbell. The weight is 30 cm from the pivot point of the elbow. What force must be exerted five from

the elbow to lift the weight? Assume everything is perpendicular.

Physics
1 answer:
qwelly [4]3 years ago
4 0
Refer to the diagram shown below.

The force, F, is applied at 5 cm from the elbow.

For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N

Answer: 1500 N

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A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a hal
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d = 44.64 m

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F = ma

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a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2

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