Force Diagram

[I Love Pi]

I already solved for the five items Force of Gravity (Weight), Normal Force, Tension, Friction (kinetic), and acceleration. Also, the coefficient of kinetic friction was given in the problem. Thus, I solved for six of the seven required items and I need to solve for one more.

I need to solve for one of the following: either Applied Force, Friction (static), Elastic/Spring Force, or the coefficient of static friction. I listed my thought process of what I know about each of the items and need help determining which force I can solve for. Could you please help me determine the last item from the checklist to solve for? As always, thank you!

 

[topsquark]

I already solved for the five items Force of Gravity (Weight), Normal Force, Tension, Friction (kinetic), and acceleration. Also, the coefficient of kinetic friction was given in the problem. Thus, I solved for six of the seven required items and I need to solve for one more.

I need to solve for one of the following: either Applied Force, Friction (static), Elastic/Spring Force, or the coefficient of static friction. I listed my thought process of what I know about each of the items and need help determining which force I can solve for. Could you please help me determine the last item from the checklist to solve for? As always, thank you!

I'm confused about the setup here.

What is a "marker?" What is the "heavy one?"

-Dan

 

[I Love Pi]

The marker is resting on the table. Both the marker and Rubik’s cube are connected to the same string to create a pulley.

The marker weighs 0.016 kg while the Rubik’s cube weighs 0.1 kg. So, the marker is the “lighter object” while the Rubik’s cube is the “heavy one.”

I reattached the document and made readjustments for clarity.

Let me know if you have any other inquiries.

 

[topsquark]

The marker is resting on the table. Both the marker and Rubik’s cube are connected to the same string to create a pulley.

The marker weighs 0.016 kg while the Rubik’s cube weighs 0.1 kg. So, the marker is the “lighter object” while the Rubik’s cube is the “heavy one.”

I reattached the document and made readjustments for clarity.

Let me know if you have any other inquiries.

Thank you.

Okay, first, I have an issue with the problem as stated because it's asking for things that take place at two different points in time and doesn't explicitly tell you that. (I'm guessing.) For example, I'm taking the "applied force" to be a force applied to the system before the cube is dropped. But in that case there is no acceleration, which must come after the cube is dropped.

You are perhaps not counting enough quantities you can calculate.
Before release: N, mg, Mg, T, F
After release, additionally: T, fk, a.
(The two tensions will be different.) The other quantities do not exist for this problem.

-Dan

 

[I Love Pi]

Okay, based on you advise, I solved for as much as I could. It appears I cannot solve for an applied force since the hand does not apply a push/pull. I am not entirely sure if I still need to solve for another tension, as you mentioned the tensions would be different. It was to my knowledge that tensions in a pulley system are of equal value and opposite magnitude (Newton’s 3rd Law).

 

[Deleted member 4993]

as you mentioned the tensions would be different.

The different tensions are

• the tension before release - no acceleration
  
  
• tension after release - with acceleration

 

[topsquark]

Okay, based on you advise, I solved for as much as I could. It appears I cannot solve for an applied force since the hand does not apply a push/pull. I am not entirely sure if I still need to solve for another tension, as you mentioned the tensions would be different. It was to my knowledge that tensions in a pulley system are of equal value and opposite magnitude (Newton’s 3rd Law).

This is why I'm suggesting that this is a badly worded problem. Before the cube is released I am assuming that there is a force acting on the marker to hold it in place. That's what I'm calling the applied force. We can calculate the tension in the string based on this assumption. When the cube is released we take away the applied force and the system starts to accelerate and, as the marker starts to slide we get friction. The tension will be different then.

-Dan

 

[I Love Pi]

Okay. I solved for all of the variables before the marker is released; without acceleration and after the marker is realeased; with acceleration. I found the tension before and after the marker is released. Also, I found the tension before the marker is released to be the same force as the applied force of the hand holding the marker from accelerating. Thank you for your help! You are awesome!

 

[Deleted member 4993]

Okay. I solved for all of the variables before the marker is released; without acceleration and after the marker is realeased; with acceleration. I found the tension before and after the marker is released. Also, I found the tension before the marker is released to be the same force as the applied force of the hand holding the marker from accelerating. Thank you for your help! You are awesome!

I do not see any friction force (effect of static friction prior to motion and effect of dynamic friction after motion) - those need to be included!

 

[I Love Pi]

I do not know how to find the static friction (the coefficient of static friction was not given). How would I solve for static friction?

Also, I did not learn how to find dynamic friction?

 

[Deleted member 4993]

I do not know how to find the static friction (the coefficient of static friction was not given). How would I solve for static friction?

Also, I did not learn how to find dynamic friction?

But you need to include those in your free-body-diagram -

friction force = µ * N​

 

[topsquark]

But you need to include those in your free-body-diagram -

friction force = µ * N​

I didn't look over the actual numbers but the rest looks good. I wouldn't include a static friction force (though one clearly does exist) because then we wouldn't be able to figure out an applied force. What's holding the marker can't solely be static friction because when we "release" the marker it starts to slide. Again, it's a difficulty with the problem statement but I'd simply ignore static friction here.

As to the kinetic friction I think his latest diagram does just fine. What is it that bothers you about it? Maybe I'm missing something. (All too possible!)

-Dan