Mr.p: Good morning. Billy, please read the problem and Bobby, please translate.
"Flipping Physics" Intro
Bobby: To tighten a lag bolt, a...
Bobby: Hold up Billy. What is a lag bolt?
Mr.p: A lag bolt or a lag screw has a screw on one end and an hexagonal nut on the other end.
The one in this example looks like this.
Bobby: Okay, Billy?
Billy: To tighten a lag bolt,
a 29N force is applied at a 90 degree angle to a wrench 0.18 meters from the center of a lag bolt.
Bobby: Force equals 29 newtons. Theta equals 90 degrees.
And the distance, d, from the center of the lag bolt to the force is 0.18m.
Billy: If the angle between the wrench and the force is changed to 50 degrees,
what magnitude force is necessary to tighten the bolt with the same torque?
Assume both angles have two significant digits.
Bobby: Theta equals 50 degrees. Force equals question mark. Torque equals torque.
And the angles have 2 significant figures.
Mr.p: This is a fine start, however, it needs a little lovin'. Billy, or Bo, could you please help?
Bo: Well, there are two different times a force is applied to the wrench.
So we should label those variables with 1's and 2's.
Billy: Right. So 29 newtons is force 1. 90 degrees is theta 1. 50 degrees is theta 2.
Both torques are equal, so torque 1 equals torque 2. And the force we are solving for is force 2.
Bobby: And the distance is actually not d, it is the r vector, because we are solving for torque and
torque equals r, the distance from the axis of rotation to where the force is applied, times the force,
times the sine of theta, the angle between the direction of the force of the direction of the r vector.
So r1 equals r2 and both equal 0.18 meters.
Mr.p: Can anyone tell me what is missing from the torque equation as currently shown?
Bobby: Vector symbols. Torque is a vector. So there needs to be vector symbols over tau, r, and F.
Billy: But the question clearly states we are looking for magnitude. So, do we even need the vector symbols?
Mr.p: Billy, you are correct that we are only solving for the magnitude of the force of this problem,
however, it is a good idea to remember torque is a vector. Therefore, we are going to add
the vector symbols and then indicate we are going to disregard them because we are solving for magnitude.
Bo, could you please solve the problem?
Bo: Well torque 1 equals r 1 times force 1 times sine of theta 1 or 0.18 times 29 times sine of 90,
which equals 5.22 newton meters.
Because torque 1 and torque 2 are equal, torque 2 also equals 5.22 newton meters.
We can rearrange that equation for torque 2 to solve for force 2, which then equals
torque 1 divided by the quantity r 2 times sine of theta 2. With numbers that is...
Billy: Just a second Bo.
Theta 2 is not 50 degrees. The angle between r vector 2 and force 2 is actually 130 degrees.
Bo: Why?
Billy: Well r vector 2 points to the right and force 2 points down and to the left.
The angle between those two directions is going to be 130 degrees not 50 degrees.
You can even see the way mr.p drew the illustration shows theta 2 cannot be 50 degrees.
Bobby: I agree with that, but it does not really matter.
Bo: Why not?
Bobby: 130 and 50 degrees are supplementary angles and the signs of supplementary angles are identical.
Bo: What?
Bobby: Look at the shape of the sine curve.
The sine of 90 + x
and the sine of 90 - x give the same value.
And 90 + x and 90 - x are supplementary angles,
they add up to 180 degrees.
Bo: Okay. So we could really use 50 or 130 degrees.
Billy: Regardless, 130 degrees is the correct angle to use.
Bo: I feel like we've had this conversation before
Billy: It's 130.
Bo: Let's plug in numbers.
5.22 divided by the quantity 0.18 times the sine of 130 degrees equals 37.8568 or 38 N with 2 significant digits.
Mr.p: Now watch as I pull at the new angle.
You can see we get roughly 38 newtons.
It does not work out perfectly, likely because the friction between the wood and the screw is not consistent.
Okay, let's add a part B to the problem. Bobby, please read and Bo, please translate.
Bobby: If a pipe is fitted to the wrench which
increases the distance between the lag bolts
and where the force is applied to 1.08 meters,
what would the minimum magnitude force necessary be to cause the same torque as before?
Bo: This is a third torque which is the same as torque 1. So force 3 equals torque 1 divided by r 3,
which is 1.08 meters times sine of theta 3.
We know an angle of 90 degrees provides the largest torque because the sine of 90 equals 1.
Therefore, because we are looking for the minimum force, theta 3 needs to be 90 degrees.
So force 3 equals 4.83 with repeating 3 which rounds to 4.8 newtons with two significant digits.
Bobby: Nice translating.
Bo: Oh yeah. Sorry. I was just supposed to translate.
Mr.p: It's okay Bo. You do not need to apologize for solving the problem.
I have added a pipe to the wrench and
you can see when I pull at a 90 degree angle 1.08 meters
the lag bolt, we do get roughly a 4.8 newton force.
Again it does not work out perfectly
because of the inconsistencies of the friction between the wood and the screw.
But now, I actually also had to hold up the pipe which I'm sure added some unknown forces to the pipe.
It still does end up being pretty close though. Just so you know, typically mechanics do carry
a metal pipe in their toolbox for situations where they need more torque to loosen or tighten nuts and bolts.
They call it a "cheater pipe" because they can use the pipe to extend the length of their wrenches.
This increases the length of the lever arm
and therefore increases the amount of torque they can apply using the wrench.
When using a cheater pipe, you do have to be careful though
because you may end up causing an angular acceleration of the wrench instead of the bolt.
Which is exactly what happened to this wrench.
Too much torque was applied using a cheater pipe and the wrench was angularly accelerated
and permanently bent.
Thank you very much for learning with me today, I enjoyed learning with you.