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 is at an angle if 30 degrees to the rod and causes a cw torque. what is the net torque about the pivot?
can someone help me out im totally lost on this and don't even know where to start.
Is there any reason why you would not sum torques about the piviot point?
Remember torque=force*distance*sinTheta
i got 15 is that correct?
To find the net torque about the pivot point, we need to calculate the torque caused by each force and then add them together.
The formula for torque is T = F * r * sin(theta), where T is the torque, F is the force, r is the distance from the pivot point, and theta is the angle between the force and the lever arm.
Let's calculate the torque caused by the first force of 6.0N applied perpendicularly at a distance of 1.2m from the pivot. Since the force is perpendicular, the angle theta is 90 degrees, and sin(90) = 1.
Therefore, the torque caused by the first force is T1 = 6.0N * 1.2m * 1 = 7.2 Nm (counterclockwise).
Now, let's calculate the torque caused by the second force of 5.2N at the end of the rod, 3.0m from the pivot, at an angle of 30 degrees to the rod. The angle theta is 30 degrees, so we need to calculate sin(30).
Using a scientific calculator or reference tables, we find that sin(30) ≈ 0.5.
Therefore, the torque caused by the second force is T2 = 5.2N * 3.0m * 0.5 = 7.8 Nm (clockwise).
To find the net torque about the pivot point, we can subtract the clockwise torque from the counterclockwise torque.
Net torque = T1 - T2 = 7.2 Nm - 7.8 Nm = -0.6 Nm
The net torque about the pivot is -0.6 Nm. The negative sign indicates a clockwise rotation.