# Inclined plane FBD for latch mechanism

• JoeyF
In summary, The designer is trying to form a Facebook group for a mechanism that uses linear motion to actuate a pin in a transverse direction. They have attached a general idea of the mechanism, along with a more detailed FBD representing actual CAD. They would like to know the relationship between the incline angle, and resultant push force required to move the body. When they plug their systems of equations into Wolframalpha, they get negative results for N2 and Fp, which indicates they are making wrong assumptions. They even tried a couple different coordinate systems with no luck. Their hunch is that their assumption highlighted in red is incorrect. They then have a classic wedge problem. Assuming all surfaces are
JoeyF
TL;DR Summary
Having trouble forming a FBD to understand a latch mechanism I'm designing. See details below.
I apologize if this is the wrong area to post this in, I've never posted on thisforum before.

I'm trying to form a FBD of a mechanism that uses linear motion to actuate a pin in a transverse direction. I've attached the general idea in a picture where the pin is free to move up and down. In reality the pin is more of a spring, but I think it will be easier to model as a pin with a given weight so I can assume constant force in that direction. Also attached a more detailed FBD representing actual CAD.

I would like to know the relationship between the incline angle, and resultant push force required to move the body. When I plug my systems of equations into Wolframalpha, I get negative results for N2 and Fp, which indicates I am making wrong assumptions. I even tried a couple different coordinate systems with no luck. My hunch is that my assumption highlighted in red is incorrect.

This is my work so far. Thanks in advance!

General Idea:

FBD for Purple Body:

Assumptions:
• Frictionless Top surface of purple Plunger (green surface)
• Normal Force in X direction is equal to Fs (Fsx = N1)
Known Variables:

Mu1 = Mu2 = 0.3

Fs = 17N

Unknown Variables:

Fp =? (Plunger Push Force)

N2 =? (normal force from cap)

Declaration of Equations:

Fsx = Fs*cos(theta)

N1 = Fsx

Ff1 = N1*Mu1 …. N1 = Ff1/Mu

Fsy = Fs*sin(theta)

Ff2 = N2*Mu2Summation of Forces:

Sum of Forces in X = N1 – Fsx + N2*cos(theta) - Ff2*sin(theta) + Fp*sin(theta) --------->

0 = 17*cos(theta) – 17*cos(theta) + N2*cos(theta) - N2*0.3*sin(theta) + Fp*sin(theta)

Sum of Forces in Y = Ff1 – Fsy – N2*sin(theta) + Ff2*cos(theta) – Fp*cos(theta) --------->

0 = 17*cos(theta)*0.3 – 17*sin(theta) + N2*sin(theta) + N2*0.3*cos(theta) - Fp*cos(theta)

I would flip your diagram so “motion in” is in the +x direction, so that “motion out” will be in the +y direction.
Resolve forces at the contact of the pin on the wedge, to be in the x and y directions.
You then have a classic wedge problem.

Is the pin restrained to remain vertical? If so, how?

JoeyF and hutchphd
I agree that the biggest problem is how you hold the pin and that will be the important design element
to get a feel, first assume all the surfaces are frictionless. Then the "force multiplier" is the inverse of the "distance multiplier"##=\ tan (\theta)##. So a slope of a half will double the force delivered to the pin. Friction will change it a little but the design intent will be to minimize friction

JoeyF
Thanks Guys,

I think I found my main error. I had the Normal force on the plunger body backwards. This should have been pointing inward to the body rather than outward. I did however, redo my coordinate system like you suggested, split the bodies, and added a roller assumption to the vertical pin. Cleans things up nicely! Let me know if you see anything wrong with it, however, I'm getting values that make sense finally.

I did notice when I plug in too large of incline angles, the resultant forces end up switching negative. I assume this is the point in which the wedge "self locks" ? It's greater than 65 degrees or so.

JoeyF said:
I did notice when I plug in too large of incline angles, the resultant forces end up switching negative. I assume this is the point in which the wedge "self locks" ? It's greater than 65 degrees or so.
Fundamentally, a wedge or taper is “self locking” when the arctangent(taper angle) is less than the friction coefficient. But I thought you had eliminated friction in your model.

JoeyF
This is kind of what I was after. In order to move the button 2.8 mm, these are the distances needed with forces to go along with the wedge angle.

In a friction free world, the work done on the wedge, is also the work done on the pin.
Work done is force * distance moved.
At 45° the ratio is 1:1, and I read a wedge force of 28 units.
That should be the same as the vertical force on the pin. Is that the case?

Baluncore said:
In a friction free world, the work done on the wedge, is also the work done on the pin.
Work done is force * distance moved.
At 45° the ratio is 1:1, and I read a wedge force of 28 units.
That should be the same as the vertical force on the pin. Is that the case?
It may be the case, but my graph is showing the output assuming friction.
In either case, my model has the Normal force to both the pin and the plunger at contact being equal. And if it was a 45° incline, then I would expect the travel to be the same, so that would be equal amounts of work done on both bodies.

JoeyF said:
And if it was a 45° incline, then I would expect the travel to be the same, so that would be equal amounts of work done on both bodies.
Yes, and the position graph shows about 2.8 units of wedge movement at 45°.
I see no way of knowing what friction coefficient you used for that graph, nor what force was on the pin. Can you reveal those numbers?

Baluncore said:
Yes, and the position graph shows about 2.8 units of wedge movement at 45°.
I see no way of knowing what friction coefficient you used for that graph, nor what force was on the pin. Can you reveal those numbers?
Yep, these are the equations I plugged into Wolfram. I varied the theta value and it spit out my N1 variable as well as the Fp variable (push force). A force of 17 N is the external force on the pin. X is my friction coefficient (0.3).

N*X*cos(theta)+N*sin(theta)+X(N*X*sin(theta)-N*cos(theta)) = F,

N = 17/(cos(theta)-X*(theta)),

X = 0.3,

theta = 25*pi/180

## 1. What is an inclined plane FBD for latch mechanism?

An inclined plane FBD (free body diagram) for latch mechanism is a visual representation of the forces acting on a latch mechanism that is placed on an inclined plane. It helps to analyze the forces involved and understand the equilibrium of the mechanism.

## 2. How is an inclined plane FBD for latch mechanism drawn?

An inclined plane FBD for latch mechanism is drawn by first identifying all the forces acting on the mechanism, such as the weight of the latch, the normal force, and the force of friction. These forces are then represented as arrows on the diagram, with their direction and magnitude labeled.

## 3. Why is an inclined plane FBD important for latch mechanisms?

An inclined plane FBD is important for latch mechanisms because it allows us to understand the forces involved in keeping the latch in place on an inclined plane. It also helps in determining the minimum force required to move the latch and the direction in which the force should be applied.

## 4. How does the angle of inclination affect the forces on a latch mechanism?

The angle of inclination affects the forces on a latch mechanism by changing the direction and magnitude of the forces. As the angle increases, the normal force decreases while the force of friction increases. This affects the equilibrium of the mechanism and the amount of force needed to move the latch.

## 5. Can an inclined plane FBD be used for other types of mechanisms?

Yes, an inclined plane FBD can be used for other types of mechanisms as well. It is a useful tool for analyzing the forces acting on any object on an inclined plane, not just latch mechanisms. It can be used for objects like ramps, slides, and even vehicles on inclined roads.

• Mechanics
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
3K
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
3K
• Introductory Physics Homework Help
Replies
19
Views
3K
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
5K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
21
Views
3K