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An iron bolt of mass 57.5 g hangs from a string 38.8 cm long. The top end the right from the previously vertical line f the string is fixed. Without touching it, a magnet attracts the bolt so that it remains stationary, but is displaced horizontally 22.5 cm to the string (a) Draw a free-body diagram f the bolt. (b) Find the tension the string. (c) Find the magnetic force on the bolt.

Respuesta :

Answer:

tension is 0.692 N

magnetic force is 0.401 N

Explanation:

given data

mass = 57.5 g = 0.0575 kg

long = 38.8 cm = 0.388 m

displace = 22.5 cm = 0.225 m

to find out

free-body diagram ,  tension and magnetic force

solution

we consider here angle is θ

so free body diagram is attach below

and

here cosθ = x/l

so θ = cos^-1 ( 0.225 / 0.388)

θ = 54.6°

so tension from net  force is

net vertical force

t sinθ - F(g) = 0 ...........1

t = F(g) / sinθ

t = mg /sin54.6

t = 0.0575 ( 9.8) / sin54.6

t = 0.692

so tension is 0.692 N

and

net horizontal force

- t cosθ + F(magnetic) = 0   ....................2

t cosθ = F(magnetic)

put here all value

F(magnetic) = 0.692 × cos 54.6

F(magnetic) = 0.401

so magnetic force is 0.401 N

Ver imagen DeniceSandidge

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