Question
A force of 32 N is applied at the end of a wrench handle that is 27 cm long. The force is applied in a direction perpendicular to the handle as in the diagram.
(a) What is the torque applied to the nut by the wrench?
(b) What would the torque be if the force were applied half way up the handle instead of at the end?
(a) What is the torque applied to the nut by the wrench?
(b) What would the torque be if the force were applied half way up the handle instead of at the end?
Answers
Anonymous
a.)N= r*F
(32N)(.27m)= 8.64 N.m
b.) half of .27=0.135
t= F*d
(32N)(0.135)= 4.32 N.m
(32N)(.27m)= 8.64 N.m
b.) half of .27=0.135
t= F*d
(32N)(0.135)= 4.32 N.m