Friction and Scale Readings for a Cuboid on a Weighing Scale

[Eitan Levy]

Homework Statement

A cuboid with a mass of M is put on a weighing scale.
First the situation is the one in the picture (the ball doesn't move), the cuboid stays on the cuboid without moving.
What will the scale show?
Now yarn number 2 is torn, the cuboid still doesn't move.
What will the scale show right after the yarn is torn?
What is the friction force that the scale will apply on the cuboid right after the yarn is torn?
What will the scale show when α=0?
What is the friction force will apply on the cuboid right when α=0?

Homework Equations

a_r=v²/r
ma=F

The Attempt at a Solution

I don't have the answers to my question so I have no idea if I am correct or not.
When yarn number 2 still isn't torn:

mg=T₁cosα
T₁sinα=T₂
So, f_s=0
N=T₁cosα+Mg=(M+m)g[/B]

Right after the yarn is torn:

a_r=v²/r=0
T₁=mgcosα
N=Mg+T₁cosα
N=Mg+mgcos²α
f_s=T₁sinα
f_s=mgsinαcosα

When α=0

f_s=0
Let's say the the length of yarn number 1 is l
Energy right after the yarn is torn: E₀=mgl(1-cosα)
Energy when α=0: E₁=0.5mv²
E₀=E₁
v²=2-2cosα
ma_r=T₁-mg
a_r=v²/r
T₁=mg(1+2-2cosα)=mg(3-2cosα)
N=Mg+T₁=Mg+mg(3-2cosα).


Does everything seem correct? I am really uncertain about this.
Thanks a lot!

 

[kuruman]

You show too many forces in your diagrams. It seems you are confused about what your system is. First you should decide what your system is. Your choices are (a) cuboid only; (b) hanging mass with string only; (c) cuboid and hanging mass together. Then consider what objects outside your chosen system act on it. Draw them in and analyze them.

Also, take a good look at your last drawing for it illustrates what I am saying. It shows an unbalanced force to the left, yet you claim the acceleration is zero. Can that be?

 

[Eitan Levy]

You show too many forces in your diagrams. It seems you are confused about what your system is. First you should decide what your system is. Your choices are (a) cuboid only; (b) hanging mass with string only; (c) cuboid and hanging mass together. Then consider what objects outside your chosen system act on it. Draw them in and analyze them.

Also, take a good look at your last drawing for it illustrates what I am saying. It shows an unbalanced force to the left, yet you claim the acceleration is zero. Can that be?

I thought it would better visually for you to look on. When I solve I draw each body by itself.
I show the force but says it equals to zero later, obviously in the exam I won't draw that force but explain why it's zero.

 

[kuruman]

I thought it would better visually for you to look on.

It is not because it isn't clear what you are drawing a free body diagram of. This way of drawing free body diagrams will lead you into trouble when you get to problems that involve torques where the point of application of the forces matters. Although your answers are all correct, there should be no forces drawn on the left side of the box because there is nothing on that side that can exert a force. The reaction forces due to the tension in the strings should be drawn at the other side of the strings where it's tied to the ceiling and the wall. Also, friction should be drawn at the point of contact.

Drawing action-reaction counterparts in a multi-component free body diagram does no harm if done correctly because one adds and subtracts a vector, i.e. adds the zero vector. However, that makes the picture more confusing to read and increases the likelihood of adding something that doesn't belong.

 

[Eitan Levy]

It is not because it isn't clear what you are drawing a free body diagram of. This way of drawing free body diagrams will lead you into trouble when you get to problems that involve torques where the point of application of the forces matters. Although your answers are all correct, there should be no forces drawn on the left side of the box because there is nothing on that side that can exert a force. The reaction forces due to the tension in the strings should be drawn at the other side of the strings where it's tied to the ceiling and the wall. Also, friction should be drawn at the point of contact.

Drawing action-reaction counterparts in a multi-component free body diagram does no harm if done correctly because one adds and subtracts a vector, i.e. adds the zero vector. However, that makes the picture more confusing to read and increases the likelihood of adding something that doesn't belong.

Alright, thank you for your advice.

 

[Billing Tom]

Looks like a good advice. Thank you so much!