What Forces Affect Motion in Block Dynamics?

[superhero241]

Hello,

I have an urgent question due soon, and I have no idea how to solve it.

 

[superhero241]

I have all the answers and they turned out to be right. I just need help with the last one.

Here is what I did:

Fk = MuK N
Fk = 0.41 (167)(9.8)
Fk = 671.006

F = ma
671.006 = 167 a
a = 4.018 m/s^2

However, it's not the right answer. Please help!

 

[berkeman]

I have all the answers and they turned out to be right. I just need help with the last one.

Here is what I did:

Fk = MuK N
Fk = 0.41 (167)(9.8)
Fk = 671.006

F = ma
671.006 = 167 a
a = 4.018 m/s^2

However, it's not the right answer. Please help!

I don't see a cos(θ) term in your equations. That is needed when calculating the effective normal force on the ramp.

 

[superhero241]

Ok so let me redo this.

Fk = MuK N
Fk = 0.41 (167)(9.8)cos52
Fk = 413.11 N

F = ma
413.11 = 167 a
a = 2.47 m/s^2

The answer is still wrong. I am pretty sure I messed up because I didn't use the M2 in part c that they give me.

 

[berkeman]

Ok so let me redo this.

Fk = MuK N
Fk = 0.41 (167)(9.8)cos52
Fk = 413.11 N

F = ma
413.11 = 167 a
a = 2.47 m/s^2

The answer is still wrong. I am pretty sure I messed up because I didn't use the M2 in part c that they give me.

You are working the first problem of the 3, correct?

The first equations where you calculate Fk are correct.

But I don't understand what you are doing in the 2nd set of equations. The force of tension up by the rope on the M2 bucket is balancing what force down on the bucket?

 

[superhero241]

I'm solving on the last problem. Sorry for the confusion.

I am using the second set of equations to solve for acceleration of M1.

 

[berkeman]

I'm solving on the last problem. Sorry for the confusion.

I am using the second set of equations to solve for acceleration of M1.

Ah, no wonder I was off in the weeds

The mass that gets accelerated (by the difference in the forces) is the sum of the two masses. Does that help?

 

[superhero241]

I don't understand. So I add the 2 masses in the first set and second set of equations?

 

[berkeman]

I'm kind of de-synchronized from what equations you have right now. Maybe re-summarize them?

In the 3rd question, you have more force than needed to break M1 loose and cause the M1 & M2 combined masses to accelerate to the right & down. The acceleration is defined by the sum of the forces (they are in opposite directions), i.e., the tension in the rope, and by the sum of the two masses, since that is what is being accelerated.

 

[superhero241]

It's giving me the coefficient of kinetic friction for M1. So I do:

Fk = Muk*N
Fk = 0.41(167)(9.8)cos52 = 413.11 N

Then, to find the acceleration, I use the F = ma equation. so you said add the two masses for m.

413.11 = (167+231.3) a
a = 1.04 m/s^2 but it's wrong

 

[berkeman]

It's giving me the coefficient of kinetic friction for M1. So I do:

Fk = Muk*N
Fk = 0.41(167)(9.8)cos52 = 413.11 N

Then, to find the acceleration, I use the F = ma equation. so you said add the two masses for m.

413.11 = (167+231.3) a
a = 1.04 m/s^2 but it's wrong

The 413.11N is the force due to friction on M1. It is not the sum of the forces acting on the two bodies together.

The better way to write the equations would be like...

$\sum F = M_{total} a$

The sum of the forces is the friction force on M1, and what force on M2?

The total mass is what? And that gives you what resulting acceleration for the system of M1 and M2 tied together?

 

[superhero241]

so ∑F = 413.11? if not, how do I find the force on M2?

 

[berkeman]

so ∑F = 413.11? if not, how do I find the force on M2?

If you draw a FBD for M2, the tension up is due to the frictional force from M1. What is the force down on M2 due to?

 

[superhero241]

Gravity.

 

[berkeman]

Ding ding ding.

So draw the FBD for both masses, and show the forces that act on the system. The imbalance in those forces is what causes the acceleration of the two masses together.

Show us your work...

 

[superhero241]

so Fgrav = mgsin52 = 231.3(9.8)sin52 = 1786.22 N

so ∑F=Mtotal*a

(1786.22-413.11) = (167 + 231.3)*a

a = 3.45 m/s^2 ? It's not recognizing my answer still

 

[superhero241]

Oh, I should subtract the two forces to get a net force?

 

[berkeman]

so Fgrav = mgsin52 = 231.3(9.8)sin52 = 1786.22 N

so ∑F=Mtotal*a

(1786.22-413.11) = (167 + 231.3)*a

a = 3.45 m/s^2 ? It's not recognizing my answer still

There is no sin() involved in the force of gravity on M2. It points straight down, directly opposing the tension in the rope due to M2's frictional force...

 

[superhero241]

This problem is so frustrating. After so many attempts, and my answer is still wrong. please tell me where I am messing up.

Fgrav = mg = 231.3(9.8) = 2266.74 N

so ∑F=Mtotal*a

(2266.74-413.11) = (167 +231.3)*a

a= 4.65 m/s^2

I am only trying to find the acceleration of M1...therefore I am only dividing by 167 kg/ Please help, I can't see where I am messing up!

 

[berkeman]

This problem is so frustrating. After so many attempts, and my answer is still wrong. please tell me where I am messing up.

Fgrav = mg = 231.3(9.8) = 2263.53 N

so ∑F=Mtotal*a

(2263.53-413.11) = (167 )*a

I am only trying to find the acceleration of M1...therefore I am only dividing by 167 kg/ Please help, I can't see where I am messing up!

231.3kg * 9.8m/s^2 does not quite equal 2263.53, according to my calculator.

And you still only have on of the two masses instead of the total mass on the RHS of your last equation.

 

[superhero241]

Ok it's a calculator error, but still not the right answer.

231.3*9.8 = 2266.74

Even if I do (2266.74)/(167+231.3) = 5.69 m/s^2

It's not the right answer still

 

[berkeman]

Ok it's a calculator error, but still not the right answer.

231.3*9.8 = 2266.74

Even if I do (2266.74)/(167+231.3) = 5.69 m/s^2

It's not the right answer still

Please go back to your post #19 and make the *two* changes I suggested to your equations. Fix the value of the M2 weight, and make sure both masses are represented in the sum of the masses.

 

[superhero241]

I edited it, and I got 4.65 m/s^2. Is that the correct answer?

 

[berkeman]

I edited it, and I got 4.65 m/s^2. Is that the correct answer?

That's what I get. What does our friendly answer computer think?

 

[superhero241]

It still says incorrect. The question is almost due and I don't know where the error is coming from.

 

[superhero241]

The question is only asking for acceleration of M1. Aren't we calculating the acceleration for both that way?

 

[berkeman]

It still says incorrect. The question is almost due and I don't know where the error is coming from.

The question is only asking for acceleration of M1. Aren't we calculating the acceleration for both that way?

Since the rope is taut, and the masses are both moving, we should be able to use the sum of the masses and the sum of the forces to figure out the motion of both (the magnitude of the acceleration for both masses is the same).

I'm not sure what we are missing. How many sig figs does this software generally want to see?

I get:

$$a = \frac{\Sigma F}{\Sigma M} = \frac{2266.74N - 413.11N}{167.0kg + 231.3kg} = 4.654 m/s^2$$

 

[superhero241]

It never specifies, but if you're close it accepts it. Is there any other way to solve it? I really need your help

 

[berkeman]

It never specifies, but if you're close it accepts it. Is there any other way to solve it? I really need your help

Can you repost a screenshot of what the screen looks like now? Maybe we are missing something in the way the question is asked...

 

[superhero241]

 

[berkeman]

Well, the 231.3 value of M2 should be enough to break the M1 block free and start it sliding. I'd thought for a moment that the answer might be a=0 for the 3rd question, if M2 were not heavy enough to start the pair sliding.

I'm not sure what we could be missing. I've asked for other Homework Helpers to take a look at this thread, to see if they see something that should be corrected. I'll keep looking at it, and hopefully you will get somebody else to check out the work. Post a reply to my post here, so that your name shows up at the end of the thread instead of mine (that will help you get others to look in).

 

[berkeman]

I just got word from one of the HH'ers, and we have been forgetting the vertical component of M1's weight in the force summation equation. As the ramp angle gets steeper, the normal force gets less, and the vertical component of the weight of M1 gets bigger.

Can you account for that vertical component of M1's weight in the force summation?

 

[Doc Al]

What you're missing is component of M1's weight parallel to the incline when finding ƩF.

 

[berkeman]

What you're missing is component of M1's weight parallel to the incline when finding ƩF.

Thanks Doc!

 

[superhero241]

so.. what to do?