The FBD Mystery: Why Isn't Normal Force Along Radial Line?

In summary, the conversation discusses the definition and calculation of the angle ##\psi## in a problem involving a cam and roller. The solutions provided use a different definition of ##\psi## than the one used in Hibbelers engineering dynamics book, leading to a discrepancy in the calculated angle. It is determined that the value of ##\psi = -72.23## degrees is correct, and the angle calculated in the solution is actually a different angle. The conversation also mentions the importance of being careful with signs in trigonometric calculations.
  • #1
simphys
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Homework Statement
The smooth surface of the vertical cam is defined in part by the curve r = (a cosθ + b). The forked rod is rotating with an angular acceleration θ”, and at angle θ the angular velocity is θ’. Determine the force the cam and the rod exert on the roller of mass M at this instant. The attached spring has a stiffness k and an unstretched length l. Given: k = 100 N/m , θ = 45 deg l = 0.1 m , θ’ = 6 rad/s, M = 2 kg , θ” = { 2 rad/s
Relevant Equations
c ylindrical
1672857173772.png

so this is what the FBD is.... but to be fair, to me this one looks as if the normal force in the direction of the radial line, yet it isn't????

1-138-768x435.png

here in the solution, it's not along the radial line, whys that???
 
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  • #2
simphys said:
to me this one looks as if the normal force in the direction of the radial line, yet it isn't????

The normal force is not in the direction of the radial line. Here I have distorted the shape of the cam into some other shape to emphasize things. In my distortion, the normal force is more vertical than the radial direction. For the shape given in the problem, the normal direction might be less vertical.

1672860691268.png
 
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  • #3
Maybe something I said in the last question mislead you. The normal force should be aligned with the cardioids center of curvature at that point. Thats not necessarily colinear with a radial line from the arms center of rotation to that point.
 
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  • #4
In the solution of post #1, ##\psi## is defined as $$\psi = \pi + \arctan\left(\frac{r \dot \theta}{\dot r}\right)$$ where I've used dots for time derivatives. Note that for this problem, the argument of the arctangent function is negative due to ##\dot r## being negative.

But, I don't think that the angle ##\psi## as shown in the diagram of the solution given in post #1 corresponds to this definition. Rather, if ##\psi## is defined by the equation above, then ##\psi## would be angle shown in red below

1672868167471.png

The equations for ##F## and ##F_N## in the solution of post #1 are correct for this red ##\psi##.

Note that the angle between the ##r## and ##\theta## directions should be a right angle, but it's not drawn very well in the diagram.
 
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  • #5
TSny said:
In the solution of post #1, ##\psi## is defined as $$\psi = \pi + \arctan\left(\frac{r \dot \theta}{\dot r}\right)$$ where I've used dots for time derivatives. Note that for this problem, the argument of the arctangent function is negative due to ##\dot r## being negative.

But, I don't think that the angle ##\psi## as shown in the diagram of the solution given in post #1 corresponds to this definition. Rather, if ##\psi## is defined by the equation above, then ##\psi## would be angle shown in red below

View attachment 319794
The equations for ##F## and ##F_N## in the solution of post #1 are correct for this red ##\psi##.

Note that the angle between the ##r## and ##\theta## directions should be a right angle, but it's not drawn very well in the diagram.
hey, thanks a lot! I guess I'll just always be checking with ##\psi##. but.. I have another question about this. Because in HIbbelers book engeneering dynamics he uses ##\psi = arctan(r/dr/d\theta)## so that is what I use, but.. the problem here is that I get -72.23 degrees which is different from the angle calculated in the solution. I know for sure that the psi angle is the reason why my answer is wrong, but I can't see why it is wrong.. numerical values: r = 0.4414, ##r/d\theta = -01414## such that ##\psi=-72.23##. I use degrees by the way, not radians
 
  • #6
FCF32B14-1018-4CA7-A1D7-2DC908C51CA4.jpeg
This is what i have done
 
  • #7
simphys said:
in HIbbelers book engeneering dynamics he uses ##\psi = arctan(r/dr/d\theta)##
With this definition of ##\psi##, the absolute value of ##\psi## represents the angle shown below
1672940188657.png
You have to be careful with signs. For the shape of your cam, ##dr## is negative when ##d\theta## is positive. As the roller moves along the cam from point ##a## to point ##b##, you can construct the right triangle ##abc## with legs along the ##\hat r## and ##\hat \theta## directions. So, ##\tan |\psi| = \frac{|r d\theta|}{|dr|}##. You can express ##F_{N,r}## and ##F_{N, \theta}## in terms of ##F_N## and the trig functions ##\sin |\psi|## and ##\cos |\psi|##.
the problem here is that I get -72.23 degrees which is different from the angle calculated in the solution
I believe your value of ##\psi = -72.23##o is correct. The angle ##\psi## calculated in the solutions is a different angle than this. In the solutions, ##\psi## is defined as $$\psi = \pi + \arctan\left(\frac{r \dot \theta}{\dot r}\right) = \pi + \arctan\left(\frac{r d\theta}{dr}\right) =180^o
+ (-72.23^o) = 107.8 ^o$$ This is the red ##\psi## that is shown in the diagram of post #4.
 
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  • #8
simphys said:
hey, thanks a lot! I guess I'll just always be checking with ##\psi##. but.. I have another question about this. Because in HIbbelers book engeneering dynamics he uses ##\psi = arctan(r/dr/d\theta)## so that is what I use, but.. the problem here is that I get -72.23 degrees which is different from the angle calculated in the solution. I know for sure that the psi angle is the reason why my answer is wrong, but I can't see why it is wrong.. numerical values: r = 0.4414, ##r/d\theta = -01414## such that ##\psi=-72.23##. I use degrees by the way, not radians

In Hibbeler, they just state that if ##\psi## is calculated to be positive, then it measured as a counterclockwise rotation ( with the same sense as ##\theta^+##). If its negative, then its measured clockwise ( ## \theta^-##).
What they did in the solution did not follow that definition explicitly, they flipped it to preserve ##\psi## as measured with the sense of ##\theta^+##

As long as you are aware of it, either works.
 
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Related to The FBD Mystery: Why Isn't Normal Force Along Radial Line?

1. What is the normal force in the context of the FBD (Free Body Diagram)?

The normal force is the perpendicular contact force exerted by a surface on an object resting on it. In a Free Body Diagram (FBD), this force is typically represented as acting perpendicular to the surface of contact.

2. Why isn't the normal force always along the radial line?

The normal force is not always along the radial line because it is defined as perpendicular to the surface of contact, not necessarily radial. The radial line is relevant in circular motion and points towards the center of the circle, but surfaces can be oriented in various ways, causing the normal force to not align radially.

3. How does the orientation of the surface affect the direction of the normal force?

The orientation of the surface determines the direction of the normal force. For instance, if an object is on an inclined plane, the normal force will be perpendicular to the plane, not vertically upward or along any radial direction. The direction of the normal force is always perpendicular to the tangent of the surface at the point of contact.

4. Can the normal force have a component along the radial direction?

Yes, the normal force can have a component along the radial direction, especially in cases involving curved surfaces or circular motion. However, this component is just a part of the overall normal force, which remains perpendicular to the surface of contact.

5. How do we resolve forces in a Free Body Diagram when the normal force isn't along the radial line?

When the normal force isn't along the radial line, we resolve the forces into components parallel and perpendicular to the surface. This involves breaking down the normal force and any other forces into their respective components using trigonometric functions, allowing us to analyze the system accurately according to Newton's laws.

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