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!!!
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 N-m
Torque 2 is -3.0 x 5.2 sin 30 = -7.8 N-m
The net torque is -0.6 N-m
-0.6
Sure, I'd be happy to help! But first, let me ask you a question: Why did the torque go to the circus? Because it wanted to be a "twisted" performer!
Now, let's get back to your question. To find the net torque about the pivot, we need to calculate the torque contributed by each force and then add them together. The torque (τ) can be calculated using the formula τ = F * r * sin(θ), where F is the force, r is the distance from the pivot, and θ is the angle between the force and the lever arm.
For the 6.0 N force applied perpendicular to the rod at a distance of 1.2 m from the pivot, the torque will be τ1 = 6.0 N * 1.2 m * sin(90°) = 7.2 Nm.
For the 5.2 N force applied at the end of the rod (3.0 m from the pivot) at an angle of 30 degrees to the rod, the torque will be τ2 = 5.2 N * 3.0 m * sin(30°) = 7.8 Nm.
To find the net torque, we add these two torques together: τnet = τ1 + τ2 = 7.2 Nm + 7.8 Nm = 15 Nm.
So, the net torque about the pivot is 15 Nm. And remember, if you ever need assistance with physics in the future, just give a shout-out to this Clown Bot!
To solve this problem, you will need to calculate the torque caused by each force and then find their sum to determine the net torque.
The torque (τ) caused by a force (F) acting at a certain distance (r) from the pivot is given by the equation:
τ = F * r * sin(θ),
where θ is the angle between the force vector and the position vector from the pivot to the point of application of the force.
1. Let's start with the first force, which is 6.0 N applied perpendicular to the rod at a distance of 1.2 m from the pivot. Since the force is perpendicular to the rod, the angle (θ) between the force vector and the position vector is 90 degrees, and sin(90°) = 1.
τ1 = 6.0 N * 1.2 m * sin(90°) = 7.2 N·m
2. Now, let's analyze the second force of 5.2 N applied at the end of the rod, 3.0 m from the pivot, at an angle of 30 degrees to the rod.
The angle (θ) for this force is 30 degrees, and we need to calculate sin(30°) to find the torque.
τ2 = 5.2 N * 3.0 m * sin(30°)
Recall that sin(30°) = 1/2, so the second torque can be calculated as:
τ2 = 5.2 N * 3.0 m * (1/2) = 7.8 N·m.
3. Finally, to find the net torque about the pivot, you need to sum the torques caused by each force:
Net Torque = τ1 - τ2
Net Torque = 7.2 N·m - 7.8 N·m = -0.6 N·m.
The negative sign indicates that the net torque is in the clockwise (cw) direction. Therefore, the net torque about the pivot is -0.6 N·m.
Add up the torques about the pivot, treating counterclockwise torques as positive and clockwise as negative.
(6.0*1.2)+(3.0*5.2*sin30)
=15
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?