# Homework Help: Need help in Mechanics asap

1. Mar 17, 2006

### joh

I'm just gonna start with a brief introduction of myself. I'm an 18 year-old highschool student from Southern Finland, a country in Scandinavia, between Sweden and Russia.
Anyway, I'm on my last year in highschool and was actually about to graduate this spring, however I hadn't completed enough courses to do that, so I'm graduating next autumn. I want to graduate asap, so I selected some courses I hadn't completed already, and one of these courses is Physics 5, it's basicly about mechanics, motion etc. etc.
Now, I'm REALLY bad at physics and math in general , and in this course I haven't got a teacher, so I need to learn this stuff without help from the school, and if I get a 4 (our grades are from 4 to 10) in this course, I will have to do it all over again

So here comes the part where I need help, and I really need to understand how this works too, because I have 6 homework assignments that I need to figure out and complete, then April 3rd, which is in almost two weeks, I have to orally explain to the teacher how I completed these assignments, how they're calculated, why etc., and then of course there's the exam.

I would be REALLY grateful if some of you guys could help me out!

I'm going to be posting on a few other boards aswell :tongue2:

I'll start with the two first assignments, I see that they're really simple basic mechanics, but there's still something I don't understand about them (I had my last physics course like 3 years ago). You have permission to laugh, I'm a newbie :rofl: and I don't really care about that anyway, so go ahead if you must :tongue2:

http://swear.1g.fi/files/phy/01.png [Broken]

What I basicly understand from this is that the Second Law of Motion is in effect and some force of 350N is pushing the wagon forward while a force of 295N is decelerating the wagon. Then we also have gravity which is 9.81 and the mass 12kg. Oh, and the Third Law of Motion is also in effect, because gravity and then the mass of the wagon is pushing against the ground.

So basicly F = 55N, m = 12kg, g = 9.81

But then how do I calculate it, what's the formula? Am I missing some other variables?

http://swear.1g.fi/files/phy/02.png [Broken]

Here Newton's Second Law of Motion is in effect again. There's two masses. m1 is accelerating while m2 is going upwards.
So basicly there's: m1 = 8kg, m2 = 6kg, g1 = 9.81, g2 = 9.81

I'm a bit confused here though. Again, how the heck do I calculate it, what's the formula? Am I missing something?

Cheers!

Last edited by a moderator: May 2, 2017
2. Mar 17, 2006

### Staff: Mentor

Apply Newton's 2nd law!
$$\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}$$

In the first problem, you have found the net force; use Newton's 2nd law to find the acceleration.

In the second problem, try applying Newton's 2nd law to each object separately and combine the equations to solve for the acceleration. (You can also treat the two masses as a single object and find the net force along its direction of motion.)

(Note: g is just the acceleration due to gravity, a constant. There's no "g1" or "g2", just g.)

3. Mar 17, 2006

### mezarashi

Welcome to Physics Forums joh. You're in quite an interesting situation, and we here are always happy to help interested people in understanding physics.

1. You've got it right. The second law does apply. The second law is in fact an equation that looks like $$F_{net} = ma$$.

Don't get confused with the gravity. With most problems like this, you have to 'break down' the forces into the x (horizontal) and y (vertical) directions. You see that gravity acts in the downward direction, meaning it is completely perpendicular to the forces pushing and pulling the cart.

All you need to do now is apply the summation of the two forces and plug it into that equation to find the acceleration!

2. This is a confusing problem to be done the first time in your head. Motion problems like these need to be tackled first by free body analysis. Physics doesn't offer a magical equation for every situation unfortunately. To be systematic, this is what I recommend. You have two equations you must set up. The first involves analyzing mass m1. Let up be considered positive and down be negative. The second law always applies, so let's start with that.

$$F =ma$$ (second law)

Looking at the forces acting on mass m1, we see that there are two. The tension of the rope pulling up and gravity down.
$$F_{net} = T + (-F_{gravity}) = m_1(-a)$$
$$F_{net} = T - m_1g = m_1(-a)$$

I have already assumed that m1 is accelerating in the downward direction.
For the second mass m2, we do something similar and get

$$F = T - F_{gravity} = m_2(a)$$
$$F = T - m_2g = m_2(a)$$

We know that because the two are connected through a rope the magnitude of the acceleration a must be the same. We have two unknowns, the tension T and the acceleration a. We have two equations though. Now let the review of solving simultaneous equations begin =)

4. Mar 17, 2006

### andrevdh

1. Newton's second law relates the acceleration that an object experiences as a result of the forces applied to it. More specific it will accelerate according to the netto force it experiences. The two forces opposes each other so the resultant of them is $350 - 295 = 55\ N$ to the right. The other forces that it experiences is its weight downward which is cancelled by the normal force from the surface pushing it upwards just so hard as its weight is pulling it downwards. We know that these two forces cancel each out from Newton's second law other since the cart is not accelerating upward or downwards. The netto force is thus only the $55\ N$ force to the right - which will determine its acceleration according to
$$a=\frac{F_{net}}{m}$$

2. The same principle holds for the two masses on the pulley. Since the two masses are connected together they are accelerating at the same rate. We can investigate each of the two masses separately with Newton's second law. Both of them will experience an upwards tension of the same magnitude (the masses pull on the rope and the rope pulls back - Newton's third law). The tensions that a massles rope experiences on its ends from the masses pulling on it need to be the same magnitude (these tensions are in opposite directions and will cancel each other out from Newton's second law $T_{net}=ma=0a=0$)

5. Mar 18, 2006

### joh

Thanks for the help guys! I think I understand #1 now, but tell me if something is wrong here:

Normal equation for force
F = ma

The net force
350N - 295N = 55N

g & Newton III cancel eachother out, so they are irrelevant in this assignment.

Here's what we got
F = 55N
m = 12kg

Equation for acceleration
a = F/m

And there we have it
a = 55N/12kg = 4.58m/s2 ~ 4.6m/s2

I still don't quite understand #2 though

The normal equation for force is F = ma, the only one of these variables we've got is mass (m1 & m2), to get the force we still need acceleration (a), how do we get it?

Now from what you've told me the acceleration for both objects is the same because they are tied together, m1 is accelerating downwards and there's two things contributing to this movement, namely gravity (9.81m/s2) and m1's mass (8.0kg), then there's one force that's decelerating it, m2's mass (6.0kg), then gravity is also slowing down m2's deceleration of m1's acceleration.

I feel like I'm so close, but I still don't quite get it :grumpy:

How do I solve this equation simultaneoulsy ?? I've always been bad with equations :tongue2:

Last edited: Mar 18, 2006
6. Mar 18, 2006

### Hootenanny

Staff Emeritus
You need to think about the forces. In the case of the 8kg mass, the tension is going to be acting in the opposite direction to the net force, so;
$$F_{net} - T= 8g \Rightarrow F_{net} = 8g + T$$

In the case of the 6kg mass, the tension is going to be acting in the same direction as the net force, therefore;
$$F_{net} + T = 6g \Rightarrow F_{net} = 6g - T$$

As the particles are connected the net force must be equal. Therefore, you can put the equations equal to each other. Can you go from here?

7. Mar 18, 2006

### Staff: Mentor

You might want to rethink this analysis, Hoot. What do you mean by "net force"? The net force on each mass is not the same.

While you can analyze this problem in terms of the net force on both masses taken together along their direction of motion (as I mention in my post), I recommend analyzing each mass separately as mezarashi showed.

I suggest that joh attempt to solve those equations simultaneously any way he can. We'll correct any mistakes.

8. Mar 18, 2006

### Hootenanny

Staff Emeritus
Ahh yes, I see my mistake now, the same one I always made in my exams . The forces would not be the same but the acceleration would be. Therefore;

$$8a = 8g + T$$
$$6a = 6g - T$$

My mechanics tutor would crucify me

Thanks Doc Al

9. Mar 18, 2006

### Staff: Mentor

Right: Both masses have the same magnitude of acceleration (but different directions).

Almost. Consider the directions of tension force and weight for each mass.

10. Mar 18, 2006

### Hootenanny

Staff Emeritus
$$F = ma$$
$$8g - T = 8a$$
$$T - 6g = 6a$$

It's all coming back to me now...:tongue2:

11. Mar 18, 2006

### andrevdh

joh
The masses are experiencing an attractive force from the earth, we call it their weight , $W_1,\ W_2$, both of these forces pull the masses downwards towards the earth. Since one of these are bigger than the other we have the same situation as with the cart. In this case the acceleration of the system (both masses and the string) can be calculated by using their combined mass in the calculation (Newton's second law) and the netto force on the system - the combined effect of the two weights.

Last edited: Mar 19, 2006
12. Mar 18, 2006

### 3trQN

What is tension?

Is it uniform through that string?

13. Mar 18, 2006

### Hootenanny

Staff Emeritus
Tension is a force exerted on a body (in this case the string) in such a way to stretch it, tension always acts towards the string and parallel to it. It is opposite to compression.

Yes, the tension is considered uniform throughout the string.

14. Mar 18, 2006

### siddharth

To elaborate on what Hootenanny said, tension is uniform in a massless string.

Last edited: Mar 18, 2006
15. Mar 18, 2006

### arildno

Not necessarily.

The sum of forces on any portion of a massless object must be zero, according to Newton's 2.law of motion and the requirement that accelerations remain finite.

16. Mar 18, 2006

### siddharth

But does this not imply that tension is uniform? If you take a small element of a massless string, then the sum of forces on both sides must be zero and as a result, the tension is uniform.

17. Mar 18, 2006

### arildno

Nope.
Consider an Atwood machine where the pulley has mass, and that the rope slung about it does not slide relative to the pulley. (I.e, there's static friction between the pulley and the rope).

Let the tension in the rope at an angle $\theta$ be given as $T(\theta)$
and let the tangent and normal vectors at that point be given as $\vec{i}_{\theta},\vec{i}_{r}$, respectively.

Consider a rope segment of length $Rd\theta$, where R is the radius of the pulley.
The equation of motion for that rope segment is:
$$T(\theta+d\theta)\vec{i}_{\theta+d\theta}-T(\theta)\vec{i}_{\theta}+N(\theta)d\theta\vec{i}_{r}-F(\theta)d\theta\vec{i}_{r}=\vec{0}$$
which can be resolved componentwise as:
$$-T(\theta)+N(\theta)=0$$
$$\frac{dT}{d\theta}-F(\theta)=0$$
where N and F are the normal and frictional forces per unit angle acting upon the rope segment from the pulley.
It is, essentially, the increase of the tension of the rope along the pulley that provides the torque which leads to the pulley's angular acceleration.

Thus, we see that the tension will be uniform throughout a massless rope, PROVIDED there is no other force than tension acting upon the rope segment in its tangential direction.

Last edited: Mar 18, 2006
18. Mar 18, 2006

### siddharth

Yeah, I understand now. Thanks for that arildno.

19. Mar 19, 2006

### joh

A guy on another physics forum told me this would be the equation for #2:

$$F_{tension} = 2g\frac{m_1m_2}{m_2+m_1} = 67 N$$

And according to my book, his answer is correct. But I still don't quite understand how the heck he came up with that equation :grumpy:
I hope someone here can break it down for me, so it will be easier to understand, because I'm still a bit confused

Thanks

20. Mar 19, 2006

### Hootenanny

Staff Emeritus
Yes, the answer is correct. Do you see where my two equations come from?

$$8g - T = 8a$$
$$T - 6g = 6a$$

21. Mar 19, 2006

### joh

Ah, now I think it's a bit clearer.

$$m_1g - T = m_1a$$
$$T - m_2g = m_2a$$

$$F_{tension} = 2g\frac{m_1m_2}{m_2+m_1} = 67 N$$

I see where they are coming from, but what I don't see is how this other guy knew to rearrange them in the way that they are in the correct answer. I'd need some more clarification on that part if possible, then I think we can move on to the next two assignments! :)

22. Mar 20, 2006

### Hootenanny

Staff Emeritus
He just used manipulation to eliminate the acceleration. However, I think that it is easier to show the process if you do it this way;

$$8g - T =8a \Rightarrow T = 8g - 8a$$
$$T - 6g = 6a \Rightarrow T = 6g + 6a$$

Now as you have two equations which are equal to T, you can equate them;

$$6a + 6g = 8g - 8a \Rightarrow 14a = 2g$$
$$a = \frac{2g}{14} = \frac{g}{7}$$

You now have an expression for a, which you can substitute back into one of the origonal formulae (it doesn't matter which);

$$T = 6g + \frac{6g}{7}$$

You know the value of g (9.81), so you can now solve for T.

Hope this helps

23. Mar 20, 2006

### andrevdh

It is not necessary to solve simultaneous equations if you follow my previous suggestion - which leads much more natural from problem 1. Consider the combined system of masses $m_1+m_2$ to experience a netto force equal to the two opposing weights of the masses. This will give one the acceleration of the system. Subsequently one can solve for the tension by applying Newton's second law to only one of the masses. These simultaneous equations should be avoided if possible.

24. Mar 20, 2006

### joh

Thanks alot, that cleared it up for me!

Now, let's move on to the next two.

http://swear.1g.fi/files/phy/03.png [Broken]

The picture is a bit unclear, y is the velocity in meters per second, x is the time in seconds.

The text doesn't specify at which points the tension should be calculated, so let's take it at say 2 seconds where the velocity is 0.75m/s and accelerating, then at 6 seconds where the velocity is 1.5m/s and constant.

$$F = ma$$

What variables we've got:
$$m = 480kg$$
$$v_1 = 0.75m/s$$
$$v_2 = 1.5m/s$$
$$t_1 = 2s$$
$$t_2 = 6s$$

Where do we go on from here??

http://swear.1g.fi/files/phy/04.png [Broken]

I'm totally out of the loop on this one. I mean it can't be just about measuring the lengths ... can it? Hmmm ...

Last edited by a moderator: May 2, 2017
25. Mar 20, 2006

### Staff: Mentor

What matters is the acceleration; a different acceleration implies a different tension. Start by calculating the acceleration throughout the motion.