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[Mentor's Note: Thread moved from the technical forums after receiving replies]

I'm just learning torque, but if I went wrong here can someone help me (or explain the textbook explanation better).

My work was:

Torque=.4m(weight of 5kg)sin(120)

which comes out to 1.73

The answer choices are

A=.49

B=.98

C=1.7

D=2

E=3.4

The answer is D, and it is apparently because "from the diagram, sin(theta)=40cm/L, so Lsin(theta)=40cm. That is, the question gave the lever arm already. (The given angle of 60 degrees is irrelevant.) Thus, from Torque=Fl, the torque is (.4m)(.5kg)(10m/s^2)=2."

Why doesn't Torque=rFsin(theta) apply? Why is the angle irrelevant?

Thank you to all who reply; this is my first post so if I make any mistakes I apologize.

I'm just learning torque, but if I went wrong here can someone help me (or explain the textbook explanation better).

My work was:

Torque=.4m(weight of 5kg)sin(120)

which comes out to 1.73

The answer choices are

A=.49

B=.98

C=1.7

D=2

E=3.4

The answer is D, and it is apparently because "from the diagram, sin(theta)=40cm/L, so Lsin(theta)=40cm. That is, the question gave the lever arm already. (The given angle of 60 degrees is irrelevant.) Thus, from Torque=Fl, the torque is (.4m)(.5kg)(10m/s^2)=2."

Why doesn't Torque=rFsin(theta) apply? Why is the angle irrelevant?

Thank you to all who reply; this is my first post so if I make any mistakes I apologize.

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