physics classHelp!!!
posted by Steph .
A 3.0 m rod is pivoted about its left end.A force of 6.0 N is applied perpendicular to the rod at a distance of 1.2 m from the pivot causing a ccw torque, and a force of 5.2 is applied at the end of the rod 3.0 m from the pivot. The 5.2 N is at an angle of 30 degrees to the rod and causes a cw torque. What is the net torque about the pivot?
I don't have a clue how to even start this question.
PLEASE HELP!!!

physics classHelp!!! 
drwls
Add up the torques about the pivot, treating counterclockwise torques as positive and clockwise as negative.

physics classHelp!!! 
Steph
torque is radius X Force sin theta
Torque 1 and 2 have angles of 90 degrees making it equal to 1.
Torque 1:0.6 (radius of 1.2) X 6.0 N
Torque 2:1.5 (radius of 3)X 5.2 N
Torque 3:1.5 (radius of 3)X 5.2 N sine 30 degrees
Torque 1 + Torque 2 + Torque 3 = Net Torque
3.6 + 7.8 + 3.9 = 15.3
Is this correct? 
physics classHelp!!! 
drwls
No, it isn't, but thanks for showing your work. Actually there are only two torques applied. I don't see why you are applying a factor of 1/2 to each radius
Torque 1 is +1.2 x 6.0 = +7.2 Nm
Torque 2 is 3.0 x 5.2 sin 30 = 7.8 Nm
The net torque is 0.6 Nm 
physics classHelp!!! 
Anonymous
(6.0*1.2)+(3.0*5.2*sin30)
=15
Respond to this Question
Similar Questions

PHYSICS
A 3.0m rod is pivoted about its left end. A force of 6.0N is applied perpendicularto the rod at a distance of 1.2m from the pivot causing a ccw torque, and a force of 5.2N is applied at the end of the rod 3.0m fromt eh pivot. The 5.2N … 
Physics
A 3.0m rod is pivoted about its left end. A force of 6.0 N is applied perpendicular to the rod at a distance of 1.2m from the pivot causing a CCW torque, and a force of 5.2N is applied at the end of the rod 3.0m from the pivot. the … 
physics
A uniform metal rod of mass 100 kg and L (2.5m) is suspended from the side of the building. On the far end of a mass of 100 kg is hung by a rope, (distance L from a building). The rod is connected to the building on the left end by … 
physics
Please help! I am so stuck on the force diagrams A uniform metal rod of mass 100 kg and L (2.5m) is suspended from the side of the building. On the far end of a mass of 100 kg is hung by a rope, (distance L from a building). The rod … 
Physics
A uniform horizontal rod of mass 2.2 kg and length 0.63 m is free to pivot about one end as shown. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass is given by I = ￼￼￼ml2/12 … 
physics
A uniform horizontal rod of mass 2.4 kg and length 0.86 m is free to pivot about one end as shown. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass is given by I = (m*l^2)/12 If … 
physics
I am studying for a physics exam and this is the pretest. There are some that I don't understand. If someone could help me get to the answer (show all the work so I can look it all over) that would be AMAZING! Thank you very much! … 
Physics
A uniform rod of length L and mass m initially at rest is struck by a horizontal force F0 a distance x below the pivot O. The rod is free to rotate about its pivot without friction. If F0 is the average value of the force and Δt … 
Physics
The left end of a 1kg, 1.0 m long uniform thin rod is attached to a pivot, about which it can rotate freely. A force F is applied at the right end of the rod at 30 degree angle to hold the rod stationary in a horizontal position. … 
Physics
A uniform rod 2m long has a force 80N applied downward at one end. Another force 6N applied downward at the other end. At what point under the rod should the pivot be placed to keep it in equillibrim?