Help with Physics w/ Cal 1 & Projectile Motion

[sb_4000]

Hi,
I think I posted this in the wrong place, so I am posting it here. I'm taking Physics w/ cal 1, and I am having problems solving a couple of question, I was hoping some one could help, I would appreciated it..

1) block of mass m1=8kg is connected over an idead (masless and frictionless) pulley to block 2, of a mass m2=4kg. The magnitude of blocks acceleration is 3 m/s^2 and the angle of incline Theta=30 deg.
a) find the coefficient of the kinetic friction between block 2 and the plane of the incline.
b) what is the magnitude of the tension in the rope.

and also this problem which is projectile motion -
2) Projectile is fired from the top of a 40m tower at an angle of 60 deg above the horizontal hits the ground at the point 100m from the base of the tower.
a) find the speed at which the stone was thrown.
b) find the speed of the stone just before it hits the ground.
c)find the time at which the projectile hits the ground.
d) find the maximum height.

for a) i got 35 m/s

Thank You.

 

[Pyrrhus]

Do a freebody diagram for the first problem, If the pulley is ideal then it has the same tension throughout it.

For the second problem consider.

Vy = 0 at max height
Y = 0 when it hits the ground.

 

[sb_4000]

do you know how to get Part a in problem 1. I am having problems setting up the equation.

 

[Pyrrhus]

Yes, i know.

Here's a hint:

Put your coordinate system with the same inclination as theta, so the normal force will have one non-zero component, while the gravity will have two non-zero components. Also consider the other two forces Tension and friction on the coordinate system.

 

[sb_4000]

Thanks for responding to my question.

In problem two part "b" Is it F= w*sin(30)= 4N..this is what I get, I am probably doing it wrong. and In part one I still don't undrestand how to solve the problem.

 

[Pyrrhus]

Draw a freebody diagram, you can use the hint i said, to solve it.

Do you know your geometry (specially trigonometry) well?

If you still don't understand, let me know.

 

[Gokul43201]

I can't seem to say this enough (as, I imagine does cyclo) ...yet it seems, no one listens : Draw the free-body diagram. That's half the work in the problem.

 

[sb_4000]

Was my answer for "b" right?
I am still working on part a.

 

[Gokul43201]

Did you draw the free body diagram ?

 

[sb_4000]

Yes I did.

I used the book as guide to draw it.
I also worked on part a, and I get 0.38 N.
is this right?

 

[Pyrrhus]

no, that's not correct.

show me your equations.

 

[sb_4000]

I also get 0.64 when I use cos(theta).

Im using the formula F=Mu_k*mg

 

[Pyrrhus]

Try to make the equations again remembering this

\sum_{i=1}^{n} \vec{F}_{i} = m \vec{a}

 

[sb_4000]

i used N=mg*cos(theta), F=mg*sin(theta)

and then mu_k = F/N and I get 0.57 is this one right?

 

[Pyrrhus]

Ok let's try this with a different approach, you're beginning to understand, but still not there.

What are the forces acting on the x-axis on the first and second block?

What are the forces acting on the y-axis on the first and second block?

 

[sb_4000]

i also did it using F_k = ma, N= mg, and F/N, and for this one i get 0.30

 

[Pyrrhus]

Please, answer the questions of my reply above.

 

[sb_4000]

for the second block, the forces acting on it are Gravity and Normal force.
for the first block g, N, T

 

[Pyrrhus]

Try answering this questions

Aren't both blocks tied together to a string?

Do you know what's an ideal pulley?

What is the normal force?

What is friction and how does it affect movement?

 

[sb_4000]

Yes both blocks are tied to a string. I do not know what the ideal pulley is. The normal force is m*a, is the the upward motion of the block. The friction I believe slows down the movement of the block.

 

[Pyrrhus]

Yes both blocks are tied to a string. I do not know what the ideal pulley is. The normal force is m*a, is the the upward motion of the block. The friction I believe slows down the movement of the block.

Yes they are, so both blocks are experimenting tension force. An ideal pulley is a frictionless and massless pulley, so the tension throughout it will be the same (further on physics you will see a pulley with mass) . The normal force is "normal" to the surface, perpendicular to the surface. It appears because of the weight of the block on the surface, so the surface reacts exerting a force which is the normal on the block. The friction slows down movement, so it goes against the movement. Also friction appears because there's small "cracks" on a suface which affects the movement.

So now i ask again

what are the forces acting on both blocks?

 

[sb_4000]

tension, Normal force, gravity.

 

[Pyrrhus]

Maybe, i misread the problem, you have a incline plane with a block on top tied to a string that goes throught an ideal pulley and there's another block hanging? right?

 

[sb_4000]

yeah, there is a block hanging and a block on a incline, they are connected with a string that goes around a pulley.

 

[Pyrrhus]

Ok understand Normal Force, The block has a weight which is the force it exerts on the incline (part of it) and the incline reacts by exerting a force on the block which is the normal.

Read carefully!

What are the forces acting on block 1?

What are the forces acting on block 2?

 

[sb_4000]

for block 1 the hanging block, the forces are T, g

for block 2 the block on the incline the forces are T, g, N

but those the hanging block matter when you are trying to find the kinetic friction of the block on the incline?

 

[Pyrrhus]

Because of the string attached to both of them, it matters.

Now that you understood which forces act, let's try using Newton's 2nd Law

This is in the vectorial equation form

\sum_{i=1}^{n} \vec{F}_{i} = m \vec{a}

This vectorial equation will give scalar equations depending on the dimensions. This system has two dimensions so only two scalar equations.

\sum F_{x} = ma_{x}

\sum F_{y} = ma_{y}

These equations you will need it to solve the system.

Now draw a free body diagram, do you know how to draw one?

 

[sb_4000]

I do this for both of the blocks?

 

[Pyrrhus]

Yes, for both of them, and then you apply the scalar equations above for X-axis and Y-axis for each block.

 

[sb_4000]

block 2(incline)

F_x = T+(-f_k)
F_y = n+ (-g)

is this right for the block on the incline?

 

[sb_4000]

for the block hanging
F_x= 0
f_y = T +(-g)

 

[Pyrrhus]

Almost there.

are you familiar with vector components?

By the way, when it means Force of Gravity (weight) it means mass x gravity.

 

[Pyrrhus]

for the block hanging
F_x= 0
f_y = T +(-g)

Read my above post, also you're forgetting something from the formula.

By the way, are you familiar with sign conventions?

 

[sb_4000]

what are the sign conventions?

 

[sb_4000]

what am I missing from the formula?

cos(theta) and sin(theta)?

 

[Pyrrhus]

In your system, when you draw the free body diagram, you must choose which way is positive in the coordinate system. The standard is right and up positive, and left and down negative, so any vectors pointing left or down will appear as negative in the scalar equations, while the vectors pointing right and up will appear as positive. This is very important.

 

[Pyrrhus]

what am I missing from the formula?

cos(theta) and sin(theta)?

On the hanging block , it's not -g, it's mg ot -mg depending on your sign convention, and you're forgetting may.

On your analysis of the block on the incline you need to use vector's components.

 

[sb_4000]

what are the vector components again?

 

[Pyrrhus]

Well, You need to study vectors and then try another attempt to solve the problem.

 

[sb_4000]

isnt the vector components like F_x = F*cos(theta) F_y = F*sin(theta)?

 

[Pyrrhus]

I'll leave the answer in vector equations. I've to go.

Incline Block:

\vec{N} + m_{2}\vec{g} + \vec{T} + \vec{F_{f}} = m_{2} \vec{a}

Hanging Block:

m_{1}\vec{g} + \vec{T} = m_{1} \vec{a}

Also Remember

F_{f} = \mu N

 

[Pyrrhus]

isnt the vector components like F_x = F*cos(theta) F_y = F*sin(theta)?

Yes they are, try to see how you can apply them, and see if you can understand the equations i left.

 

[sb_4000]

Thanks alot..Ill try to figur it out.

 

[sb_4000]

Oene question, was the second part of my question correcty, it was 4N..the magnitude of tension..thanks again

 

[Pyrrhus]

sb4000, did you figure it out?

 

[sb_4000]

Hi,
Im still having problems solving part a, but I did part b, and I got T=39.19.

 

[Pyrrhus]

I will like to see your equations for both blocks.

 

[sb_4000]

T-m1gsin(theta) = m1a
-t+m2g=m2a

-(m1gsin(theta)+m1a)+m2g=m2a

the I got the acceleration a= m2g-m1gsin(theta)/m1+m2

T=m1gsin(theta)+m1a

this what I used to get T, my teacher had showed us a similar problem..

 

[Pyrrhus]

Your equation for the hanging block is correct, but in your equation for the incline block you forgot friction. Find tension using the equation from the hanging block.