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Statics help

  • #1
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Homework Statement


There is a copy of the problem in the picture with the graph.
upload_2015-1-31_9-5-24.png

Homework Equations


M = F*d, M = Fxr,
F_x = Fcos(theta), F_y = Fsin(theta) where theta is measured off the positive X axis,

The Attempt at a Solution


I tried splitting the force vector into X and Y components and then multiplying each of those by the perpendicular distance to the corresponding axis. My understanding from in-class examples was that these two values should sum to the moment about the given point. All I know is that I get the wrong answer:
M = 1*30cos(26.57) + 4*30sin(26.57) = 80.51

According to my book, I should get 83.2. Any help is greatly appreciated!
 

Answers and Replies

  • #2
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How did you get the angle of 26.57°?
And think about the signs.
 
  • #3
SteamKing
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Homework Statement


There is a copy of the problem in the picture with the graph.

Homework Equations


M = F*d, M = Fxr,
F_x = Fcos(theta), F_y = Fsin(theta) where theta is measured off the positive X axis,

The Attempt at a Solution


I tried splitting the force vector into X and Y components and then multiplying each of those by the perpendicular distance to the corresponding axis. My understanding from in-class examples was that these two values should sum to the moment about the given point. All I know is that I get the wrong answer:
M = 1*30cos(26.57) + 4*30sin(26.57) = 80.51

According to my book, I should get 83.2. Any help is greatly appreciated!
Two questions:
1. How did you determine the angle theta?
2. Why do you assume that the two parts which you calculated for M must be added together?
Did you check the direction of the moment each component produces about point B?
 
  • #4
13
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1. Because the squares are 1x1, I counted the squares and found the triangle to have sides of length 2 and 3. Then I used inverse tangent to sole for theta. (Apparently incorrectly. Now I get theta equal to 33.69 degrees).

2. I assumed they were added together based on an in-class example. I wasn't entirely sure what the example was supposed to be showing, so I may accidentally be solving for a couple. I don't know how to check the direction of the moment produced around B by the individual components.
 
  • #5
SteamKing
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1. Because the squares are 1x1, I counted the squares and found the triangle to have sides of length 2 and 3. Then I used inverse tangent to sole for theta. (Apparently incorrectly. Now I get theta equal to 33.69 degrees).
This angle is correct.

2. I assumed they were added together based on an in-class example. I wasn't entirely sure what the example was supposed to be showing, so I may accidentally be solving for a couple. I don't know how to check the direction of the moment produced around B by the individual components.
This is where you use your mind's power of visualization. Pretend that instead of seeing F as shown in the diagram, you see the components of F. If you compare the rotation produced by the component Fy about B to the rotation produced by the component Fx about B, you should see that one rotation will be in the opposite direction from the other. Thus, the components cannot be added together. That's why you should decide which direction of rotation is positive and which is negative when starting the problem.
 
  • #6
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Aren't the X and Y components of F both positive?

I don't understand how they're in opposite directions; they're perpendicular to each other. Once I get the X and Y components of F, if I don't add them, what do I do with them?

I tried to find the perpendicular distance d between the point B and the line of action, but F*d = M didn't give me the right answer either.
 
  • #7
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Aren't the X and Y components of F both positive?
That depends on your choice of the coordinate system. Their cross-product with the forces will certainly not have the same sign for both.

Draw the lines going through the forces: one will pass the origin at the "upper left left" side, one at the "lower right" side.
 
  • #8
SteamKing
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Aren't the X and Y components of F both positive?

I don't understand how they're in opposite directions; they're perpendicular to each other. Once I get the X and Y components of F, if I don't add them, what do I do with them?
Why don't you cut out a square like what is shown in the diagram and perform an experiment? If you hold the square at point B and tug on it at point A in the direction of each force component individually, does the square want to rotate in the same direction each time?

I tried to find the perpendicular distance d between the point B and the line of action, but F*d = M didn't give me the right answer either.
Show your calculations.
 
  • #9
13
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So the vector pointing in the first quadrant no longer means the signs are always positive - I'll have to check the direction of rotation?

The slope of the line of action is m = 2/3. Using the point (1,4), I solved for the equation of the line, which I got to be y = (2/3)x + (10/3).

The slope of d is m_d = -3/2. Because d intersects the origin in this case (I defined point B to be at the origin), the y intercept is zero, and the equation representing the line d is y = (-3/2)x.

The two lines intersect where x (and/or) y are equal, so:
(-3/2)x = (2/3)x + (10/3)
(-13/6)x = (10/3)
x = -1.54
Using one equation to sole for y and the other to check the value,
y = 2.31.

d is equal to ((-1.54)^2 + (2.31)^2)^(1/2) = 2.78, so
M = 2.78 * 30 = 83.4.

This is close to the answer, but the tolerance range for my answer to count as correct is plus or minus one in the 3rd significant digit. I'm not sure at this point if my answer is wrong because I messed up my sig figs somewhere or if it's because my trig is incorrect.
 
  • #10
SteamKing
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So the vector pointing in the first quadrant no longer means the signs are always positive - I'll have to check the direction of rotation?
When dealing with moments, always check the direction of rotation.

The slope of the line of action is m = 2/3. Using the point (1,4), I solved for the equation of the line, which I got to be y = (2/3)x + (10/3).

The slope of d is m_d = -3/2. Because d intersects the origin in this case (I defined point B to be at the origin), the y intercept is zero, and the equation representing the line d is y = (-3/2)x.

The two lines intersect where x (and/or) y are equal, so:
(-3/2)x = (2/3)x + (10/3)
(-13/6)x = (10/3)
x = -1.54
Using one equation to sole for y and the other to check the value,
y = 2.31.

d is equal to ((-1.54)^2 + (2.31)^2)^(1/2) = 2.78, so
M = 2.78 * 30 = 83.4.

This is close to the answer, but the tolerance range for my answer to count as correct is plus or minus one in the 3rd significant digit. I'm not sure at this point if my answer is wrong because I messed up my sig figs somewhere or if it's because my trig is incorrect.
This calculation is OK. If your sig figs need to be accurate to the third digit, you should probably calculate carrying four digits, and then round the final result to three digits.

Also, you should get in the habit of showing the units of the final answer and any intermediate calculations to avoid confusion.
 

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