Question

A machinist is using a wrench to loosen a nut. The wrench is 25.0 {\rm cm} long, and he exerts a 17.0-{\rm N} force at the end of the handle at 37 ^\circ with the handle


What is the magnitude of the torque does the machinist exert about the center of the nut?

What is the direction of the torque in part (A).

What is the maximum torque he could exert with this force?
How should the force mentioned in part (C) be oriented ?

Answer 1

torque = force* perpendicular distance from the axis of rotation

Answer 2

\[\tau=Nm?\]

Answer 3

\[\tau= force* r\] where r is the perpendicular distance from the axis of rotation

Answer 4

i got the answer . but whats the direction?

Answer 5

direction is upwards

Answer 6

thank you :)

Answer 7

how to get the maximum torque?

Answer 8

maximum torque can be applied if the direction of force is perpendicular to the axis of rotation

Answer 9

clear?

Answer 10

i still dont understand :( because i dont know how to get the maximum value. looking at the torque

Answer 11

well torque= force cos (theta)* radius ..now maximum value of cos (theta)=1

Answer 12

oh . so ill just equate using that equation ? to get the maximum ?

Answer 13

i got the wrong answer -_-

Answer 14

yes you need to find maximum value of the component of force for which you ca have maximum torque

Answer 15

*can

Answer 16

-_- oh my god im so confuse.

Answer 17

you can differentiate to get the maximum value

Answer 18

i just need this last question ill try to answer it though -_-

Answer 19

okay ..sorry..for now..

Answer 20

its okay :)

Answer 21

What is the maximum torque he could exert with this force?
this question

Answer 22

Maximum torque will occur if F is perpendicular to r.
\[\tau=Fr\sin\theta \] (\(\tau \) maxed if theta = 90 deg)
So the maximum torque is
max_torque=17x25xsin90
max_torque=17x25