Question

A 700 N force is applied to a lever with length 0,050 m. (a) If the force is applied at the end of the lever perpendicularly, how much torque is applied? (b) Suppose this force is applied beneath the lever arm, at an angle of 30. How much torque is applied now?

Answer 1

a.)

Torque is given by,

= F*R*sinA

here, F = 700 N

R = lever length = 0.050 m

A = angle between force and lever arm = 90 deg

then, = 700*0.05*sin(90 deg)

= 35 N*m

b.)

here, A = angle between force and lever arm = 30 deg

then, new applied torque will be:

' = 700*0.050*sin(30 deg)

' = 17.5 N*m

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