# Finding the magnitude of the moment on a pipe wrench

• yazz912
In summary, the conversation discusses finding the magnitude of the moment about point O by evaluating || OA x F ||, using vector components and the cross product. The final answer is 42sqrt(3) and the correct angle for maximum torque is θ=90 degrees. The problem is from an unknown book and chapter.
yazz912

## Homework Statement

A force of 56 pounds acts on the pipe wrench shown.
find the magnitude of the moment about O by evaluating || OA x F ||

|| OA x F ||

## The Attempt at a Solution

First I tried to find my vector component for Force which I got
-56 [ cos(30)i + sin(30)j ]

Since my force on wrench is 56 pounds and it is in a downward vertical I used F=-56
To which my vector for F simplified to =
-28sqrt(3)i -28j

Now I needed to find a component vector from O to A.
Which I initially thought would be
0i+0j +(3/2)k
Since 3/2 ft (18in) is how long my wrench is?

after I get my two vector components I use the cross product and find the magnitude. I just needed some guidance on my vectors for F and vector(OA)
.

You're making this problem harder than it should be.

In the x-y coordinate system shown in the diagram, what is the vector expression for F?

Similarly, what is the vector expression for the wrench handle OA?

You are trying to mix approaches by finding the components of F with respect to the wrench, instead of using the vectors for F and OA in the x-y system.

After my attempt to solve the problem I used the solutions manual and they had something completely different.
Which is why I started struggling .

So then would F be <0,-56>
And OA <0,18> ?

F is OK. OA is not. Remember, use the axes shown on the diagram as the reference. 18" is the length of the vector OA, and 30 degrees is the angle OA makes with the horizontal.

Would It still be ok to use the length of the wrench of 18in or convert to ft?
So OA should be
= 18[ cos(30)i + sin(30)j ]
= ( 9sqrt(3) , 9 ) ?

Yes, you can convert the length of OA to feet.

Your OA vector is still not correct. What is the length of OA?

18 inches

Well, fix your components for the vector OA.

OA= < 3sqrt(3) , 3/4 >

This is what I get after converting 18= 1.5 ft and using my reference angle.

Vector OA is still incorrect. The y-component is OK, but the x-component is off. Please show how you are calculating your components.

Oh I apologize I meant to write

3sqrt(3)/ 4 for my x component .

Looks better. Finish the calculation now that you have the proper vector expressions.

to find the magnitude of the cross product of OA and F

|| OA x F ||= 42sqrt(3) is what I get .

Looks good.

Thank you very much! One last question . Diagram listed shows theta in regards to the angle as the angle on the end of the wrench increases.
If I'm solving for theta I get 84sin(theta) .

How do I know when I am suppose to solve for the given angle such as the 30deg that was given as opposed to solving for theta in general?

yazz912 said:
Thank you very much! One last question . Diagram listed shows theta in regards to the angle as the angle on the end of the wrench increases.
If I'm solving for theta I get 84sin(theta) .

How do I know when I am suppose to solve for the given angle such as the 30deg that was given as opposed to solving for theta in general?

I'm afraid I don't quite follow what you are asking. Could you post the complete text of the question? That's why PF asks HW posters to follow the HW template when posting: it eliminates a lot of confusion for the HW helpers.

I had only posted part a) of the question. Here is the attachment. Sorry for any confusion.

In part c), you are supposed to find the angle θ such that the magnitude of the moment is maximized. What angle do you think that is? Did you make the plot that was asked for in part a)?

So in part a) was the answer correct 42sqrt(3) ? Or am I solving for θ instead of 30?

c) I believe the angle in which the moment is maximized would be at θ=90 torque is maximized when it is perpendicular.

And no I did not plot it, I'm not too familiar with the functions on my graphing calculator

yazz912 said:
So in part a) was the answer correct 42sqrt(3) ? Or am I solving for θ instead of 30?

This is the magnitude of the moment when θ = 60 degrees.

c) I believe the angle in which the moment is maximized would be at θ=90 torque is maximized when it is perpendicular.

Yes, that is a reasonable conclusion.

You may have to take a sheet of paper and draw a graph of moment versus θ to answer a).

de que capitulo y libro encontraste ese problema?

## What is a moment on a pipe wrench?

A moment on a pipe wrench is a measure of the turning force that is applied to the wrench when it is used to turn a pipe. It is a vector quantity, meaning it has both magnitude and direction.

## How is the moment on a pipe wrench calculated?

The moment on a pipe wrench is calculated by multiplying the force applied to the wrench by the distance from the point of rotation to the point where the force is applied. This is known as the moment arm or lever arm.

## What is the unit of measurement for the moment on a pipe wrench?

The unit of measurement for the moment on a pipe wrench is the Newton-meter (Nm) in the SI system or the foot-pound (ft-lb) in the imperial system.

## Why is it important to find the magnitude of the moment on a pipe wrench?

Knowing the magnitude of the moment on a pipe wrench is important for understanding the amount of force that is being exerted on the pipe and ensuring that the wrench is being used safely and effectively.

## What factors can affect the magnitude of the moment on a pipe wrench?

The magnitude of the moment on a pipe wrench can be affected by the force applied, the distance from the point of rotation, and the angle at which the force is applied. The length and thickness of the wrench, as well as the properties of the pipe being turned, can also impact the moment.

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