Foo - Physics. Plrease help if possible! :D

Hey guys I was wondering if anyone could help with the below problem. Again, I tried and tried and I am doing something wrong which I am not understanding... :(

Thanks to anyone who gives it a shot, I exhausted the book and my local sources.

Two blocks with mass 4kg and 8kg are connected by a string and slide down a 30* plane. The coefficient of friction between 4kg and plane is .25, between 8kg block and plane is .35.

a) Calculate Acceleration of each block
b) Calculate the tension
c) What happens if the positions of the block are reversed.

Answers:
a)2.21 m/s^2
b)2.26N
c)Prove upper slides down and lower on stays at rest

This is what I tried...

4kg block))

w-T1-fk=ma ---->fk=mu(mu-sub-k)(n) <---> fk=.25(sin30*)(m) <--->fk=.5
9.8(4)-T1-fk=4a
39.2-T1-.5=4a
-Ti+38.7=4a

8kg block)))
T1+w-fk=ma --->fk=mu(n) <---> fk=.35(sin30*)(m) fk = 1.4
T1+9.8(8)-1.4=8a
T1+77=8a

T1+77=8a
-T1+38.7=4a

115.7=12a
a=9.64m/s^2

I don't see my mistake and I tried everything. Since acceleration is wrong, I'll get the wrong value for T1 so part b) is no point in answering

Anyone??? I owe you long time(again)

Thanks!!

I used to be able to do problems like this. Alas 7 years from the textbookx has left my knowledge all but non-existant.

The way such problems are normally constructed, I'd hypothesize that one block will accelerate down the plane and one will decelerate (or not move). Typically, tricky instructors will make the behavior change when the order changes, or forces are applied together, etc. i.e. if smaller block has lower friction, it will slide down the plane. when it reaches the end of the string either the big block will start moving too, or it will stop the small block. Opposite should happen if small block starts uphill of the big block.

but I'm just guessing and only took one semester of high school physics 18 years ago.

The way such problems are normally constructed, I'd hypothesize that one block will accelerate down the plane and one will decelerate (or not move). Typically, tricky instructors will make the behavior change when the order changes, or forces are applied together, etc. i.e. if smaller block has lower friction, it will slide down the plane. when it reaches the end of the string either the big block will start moving too, or it will stop the small block. Opposite should happen if small block starts uphill of the big block.

but I'm just guessing and only took one semester of high school physics 18 years ago.

thanks :)

Since they are on the incline I would assume that both would slide down when they reach their maximum static friction

fs=mu(mu-sub-s)*N
Calculating static friction for the upper block, fs=.35(msin30)
fs=(.35)(4)
fs=1.4N...

I think that this is the max static friction force before the block moves. So it is pretty easy to move the top brick.

For the lower brick
fs=(.25)(4sin30)
fs=.5N

But since they are connected by this string(which I can assume is massless) would it be a combination to move both of them? Such as 1.9N?


I confused myself lol

You're trying to combine too many steps into one equation. Draw a picture and calculate individual forces. Try:

1. total friction under each block on flat surface
2. total friction under each block at 30-degree incline
3. total downward acceleration of each block on incline without friction
4. total acceleration of each block on incline with friction
5. tie them together.

The equation for the lower block, assuming it is the 4kg one, is (using m1 for 4kg to make things clearer):

m1*sin30*g - m1*f1*cos30*g - T = m1*a

NB. The component of gravity acting in the direction parallel to the inclined plane is: m1*sin30*g and not m1*g.

For the second block, the equation is:

m2*sin30*g - m2*f2*cos30*g + T = m2*a

Adding the two, T is eliminated and you get:

a = 1/2*(1 - (sqrt(3)/(m1+m2))*(0.25*m1 + 0.35*m2)))

Plugging in the values for m1 (4kg) and m2 (8kg), we have:

a = 2.21 m/s2 in the direction of the incline.

I did not compute the value of T, but I assume it is correct in the answer.

Awesome!!! thanks so much! I forgot the components of gravity and cos30

So basicly when it's an upward force at some sort of an incline, you would use cos(x)? and sin(x) for the horizontal forces?

As for part c) which asks if the blocks were reversed, the top block being 8kg and the lower being 4kg(reverse this) and now the bottom block is 8k and the top block is 4kg...

Why would the lower block stay at rest? It seems that the f(s)=mus(n) is fs=(.35)(cos30)(8)(9.8)=23.76N

The horizontal force of weight is, w=m(g)(sin30) which is 39.2N.

For the upper block
w+T1-fk=ma

9.8(4)(sin30)+t1-.25(4)(9.8)(cos30)=4a
T1+4.69=4a

For the lower block(8kg)
-T1+w-fk=ma

-T1+8(9.8)sin30-.35(8)(cos30)(9.8)=8a
-T1+39.2-23.76=8a
-T1+15.44=8a

Combinging
-T1+15.44=8a
T1+4.69=4a
20.13=12a
a=1.677 m/s^2

substituting for a, (T1+4.69=4(1.677) <---> T1=1.4302N

So,
23.76N+1.43N=25.19N
The weight is still w=m*sin30*9.8 which is 39.2N. Since weight is greater than the force keeping it steady, it would accelerate downward as well as the top block, No?

Thanks, I might have miscalculated something, Sorry.

Thanks again!!

I am not sure I quite understand the sense in which you have used the words "upward" and "horizontal" above. In the case we looked at, we looked at force balance in the direction of the incline. The force due to gravity can be "broken down" into two components, one acting in the direction parallel to the incline (the sin30 part) and the other is the normal force perperndicular to the incline direction (the cos30 part that gets multiplied by the friction coefficient). This is based on vector math (or trig.)



So basicly when it's an upward force at some sort of an incline, you would use cos(x)? and sin(x) for the horizontal forces?

Your calculation seems correct and the solution provided seems to be in error.


As for part c) which asks if the blocks were reversed, the top block being 8kg and the lower being 4kg(reverse this) and now the bottom block is 8k and the top block is 4kg...

Why would the lower block stay at rest? It seems that the f(s)=mus(n) is fs=(.35)(cos30)(8)(9.8)=23.76N

The horizontal force of weight is, w=m(g)(sin30) which is 39.2N.

I am not sure I quite understand the sense in which you have used the words "upward" and "horizontal" above. In the case we looked at, we looked at force balance in the direction of the incline. The force due to gravity can be "broken down" into two components, one acting in the direction parallel to the incline (the sin30 part) and the other is the normal force perperndicular to the incline direction (the cos30 part that gets multiplied by the friction coefficient). This is based on vector math (or trig.)
Your calculation seems correct and the solution provided seems to be in error.

Thanks! I have to work on the components.

Thanks for the help again!

As always we, on be halfof the homework helpers would like to thank everybody for their endless support.

Well it looks like you got the solution, I was kind of rearing to do "challenging" kinematics problem!

I always tell my students... DRAW A FREE BODY DIAGRAM. Label ALL the forces THEN write equations. Take each step seperatly, don't forget normal forces, etc.

Have fun with physics!

Remember, SOH CAH TOA. You're calculating vertical and horizontal force components to come up with the total using the Pythagorean theorem, hence cos30 and sin 30. And like the professors said, work on each body separately first to get their initial values then figure out their effect on each other.

Well it looks like you got the solution, I was kind of rearing to do "challenging" kinematics problem!

I always tell my students... DRAW A FREE BODY DIAGRAM. Label ALL the forces THEN write equations. Take each step seperatly, don't forget normal forces, etc.

Have fun with physics!

I draw them!! :o I know the sin(theta)=o/h, cos(theta)=a/h and tan(theta)=o/a

Actually I had a great teacher like 4-5 years ago that I will never forget the above
Its a story about a horse named Albert. She made us repeat this ten times until it was drilled into our head(which it is).

Take=Out/Albert(tan=o/a)
Saddle=Our/Horse(S=o/h)
Canter=Away/Happily(C=a/h)

We are moving onto Work now so I think it should ease up a bit :)

F=ma! Thats the most important,

F=ma!!!!!!!


PS: How do you like teaching? I am thinking of going that route.

Thanks again guys!

ovoleg,

I think you had most of the pieces correct in your first attempt anyway. One thing that may help in terms of sines and cosines in kinematics problems like this is to visualize "limiting cases", e.g., inclined plane angle = 0. In this instance, the gravitational component in the direction parallel to the plane would be 0, which would tell you immediately that using cos(theta) for this component would be wrong. Even when you get very adept at resolving force components, sanity checks such as these are helpful in catching errors.

Good luck with your physics pursuits.

A sphincter says what?