Understanding Net Force Calculation in Physics Problem #93?

[Rijad Hadzic]

Homework Statement

https://imgur.com/HR8ssJq
https://i.imgur.com/ZxAZ5NR.jpg (full problem, #93)

Homework Equations

The Attempt at a Solution

Look at the diagram with Fg, Fn, and Fb.

Then look two blocks above it, with the summation sin(theta)Fg - Fbcos(theta)

This makes no sense to me. How is sin(theta)Fg going to equal anything? Fg doesn't have any component on the x-axis at all.

So say theta was inbetween Fg and the dotted line. That means one component be directly on this line, while the other would be connecting to the end of Fg. That component that would be connecting to the end of Fg would be in the negative direction, not positive, so even if this is Fgsin(theta), I don't understand why its not negative (or firstly, why it even exists in the first place if Fg has no x components.)

 

[kuruman]

Please provide the full statement of the problem. Type it in if you must. The photograph you uploaded is truncated and blurred in places.

 

[Rijad Hadzic]

Please provide the full statement of the problem. Type it in if you must. The photograph you uploaded is truncated and blurred in places.

Sorry I uploaded a picture in the OP of the full problem.

I didn't think it was relevant because I was focused on the solution and why they had what they had at F net x direction, but on second thought not adding the problem was kinda dumb.

 

[haruspex]

Fg doesn't have any component on the x-axis at all.

Not sure what you mean by x-axis here. Do you mean horizontal?
The text says "along the plane", i.e. parallel to it, not horizontal. Fg and F_B do indeed have components in that direction.

 

[Rijad Hadzic]

Not sure what you mean by x-axis here. Do you mean horizontal?
The text says "along the plane", i.e. parallel to it, not horizontal. Fg and F_B do indeed have components in that direction.

Is Fg not negative though? It says its positive, but I don't see how it can be positive. If one component of fg is along that dotted line, the other one that connects to the head of the force has to be in the negative direction..

 

[haruspex]

Is Fg not negative though? It says its positive, but I don't see how it can be positive. If one component of fg is along that dotted line, the other one that connects to the head of the force has to be in the negative direction..

The text takes downplane as the positive direction for that axis. It is a matter of choice, as long as you are consistent. The component of F_B is up the plane so has opposite sign in the sum.

 

[Rijad Hadzic]

The text takes downplane as the positive direction for that axis. It is a matter of choice, as long as you are consistent. The component of F_B is up the plane so has opposite sign in the sum.

Ok think I'm starting to understand.

So since Fgsin(theta) points in the negative direction, and Fbcos(theta) points in the negative direction, we can say Fbcos(theta) = -Fbcos(theta) and they both have the same magnitude,

Fgsin(theta) - (-Fbcos(theta)) = 0

is my interpretation right now?

 

[haruspex]

So since Fgsin(theta) points in the negative direction, and Fbcos(theta) points in the negative direction

No, they point in opposite directions. Fg has a component down the plane while F_B has a component up the plane.

 

[Rijad Hadzic]

No, they point in opposite directions. Fg has a component down the plane while F_B has a component up the plane.

https://i.imgur.com/xS9VHXb.jpg

So this diagram must be wrong then?

Theta would actually be to the left of Fg then right. Because Fg is going down on an angle parallel with the surface?

 

[Rijad Hadzic]

It would actually look like this:

https://imgur.com/a/tq8yi

Right?

 

[haruspex]

https://i.imgur.com/xS9VHXb.jpg

So this diagram must be wrong then?

Theta would actually be to the left of Fg then right. Because Fg is going down on an angle parallel with the surface?

Your diagram marks two complementary angles as both theta. That would only be true if theta is 45 degrees.

 

[haruspex]

It would actually look like this:

https://imgur.com/a/tq8yi

Right?

That diagram is uninterpretable. You have labelled several forces as "component" without indicating which component of which force.

Fg acts straight down.
If we resolve it normal and parallel to the plane then it has a component Fg sin(θ) down the plane and a component Fg cos(θ) into the plane.
Since downplane is taken as positive by the textbook author, the component parallel to the plane is Fg sin(θ), not -Fg sin(θ).

F_B acts to the left.
If we resolve it normal and parallel to the plane then it has a component F_B cos(θ) up the plane and a component F_B sin(θ) into the plane.
Since downplane is taken as positive by the textbook author, the component parallel to the plane is -F_B cos(θ), not F_B cos(θ).

The sum of the forces parallel to the plane is therefore Fg sin(θ)-F_B cos(θ). Since it is in equilibrium, that sum is zero, so Fg sin(θ)=F_B cos(θ).

 

[Rijad Hadzic]

That diagram is uninterpretable. You have labelled several forces as "component" without indicating which component of which force.

Fg acts straight down.
If we resolve it normal and parallel to the plane then it has a component Fg sin(θ) down the plane and a component Fg cos(θ) into the plane.
Since downplane is taken as positive by the textbook author, the component parallel to the plane is Fg sin(θ), not -Fg sin(θ).

F_B acts to the left.
If we resolve it normal and parallel to the plane then it has a component F_B cos(θ) up the plane and a component F_B sin(θ) into the plane.
Since downplane is taken as positive by the textbook author, the component parallel to the plane is -F_B cos(θ), not F_B cos(θ).

The sum of the forces parallel to the plane is therefore Fg sin(θ)-F_B cos(θ). Since it is in equilibrium, that sum is zero, so Fg sin(θ)=F_B cos(θ).

I see. Well thanks for breaking it down.

https://imgur.com/a/pgCqL

Are those 2 equations at the bottom and the diagram correct now??

 

[haruspex]

I see. Well thanks for breaking it down.

https://imgur.com/a/pgCqL

Are those 2 equations at the bottom and the diagram correct now??

That diagram and equations look right.

 

[Dipti]

You need break the forces into x and y-axis of the plane and apply equilibrium equation. You will get two variables and two equations.

Follow a single sign convention. You problem of negative force will be resolved. Don't try to mix two sign conventions