Finding mass of a block tethered to an incined block above

[zombiengineer]

Homework Statement

Two blocks are connected by a string (through a frictionless massless pully) block 1 is on an inclined plane, while block two is hanging from the other end of the string.
m1 = 20kg, theta =53.1, friction between the block and the inclinde is \muk =.4
what is m2 such that it must be decended 12m in 3 s after the system is released from rest?

Homework Equations

sumF= ma
w=mg

The Attempt at a Solution

block 1 = sumF sub x =Ft-Fuk-Fgcos53.1
"""""""" = sumF sub y = Fn-Fgsin53.1=0

block 2 = sumF= Ft-Fg = ma

assuming Ft sub1 = Ft sub 2, due to the frictionless pully

Fgcos53.1+Fuk=Fg+ma

28kg*a=m*a+Fg ?

is this right so far?
I have no clue where to go from here :S

 

[zombiengineer]

Does this help at all?

I'm pretty sure I'm going in the right direction, I just don't know what to do next.

https://webmail.unm.edu/Session/36232-acFPEU5bQ2jidPBk2CKY-kmbdsse/MessagePart/INBOX/970-01-B/0927011758.jpg

 

[PhanthomJay]

Homework Statement

Two blocks are connected by a string (through a frictionless massless pully) block 1 is on an inclined plane, while block two is hanging from the other end of the string.
m1 = 20kg, theta =53.1, friction between the block and the inclinde is \muk =.4
what is m2 such that it must be decended 12m in 3 s after the system is released from rest?

Homework Equations

sumF= ma
w=mg

The Attempt at a Solution

block 1 = sumF sub x =Ft-Fuk-Fgcos53.1

Ft acts up the plane, Fuk acts down the plane, and component of the weight acts down the plane, but it is not Fgcos53, you have a trig error. Also, you must ultimately calculate Fg and Fuk.

"""""""" = sumF sub y = Fn-Fgsin53.1=0

Trig error in the weight component perpendicular to the plane.

block 2 = sumF= Ft-Fg = ma

Yes, and you can determine 'a' from the kinematic equations. m is m2.

assuming Ft sub1 = Ft sub 2, due to the frictionless pully

yes

Fgcos53.1+Fuk=Fg+ma

I think you mean Ft and not Fg after the equal sign. Also watch subscripts, it is m2. also, correct that Fgcos53 term as mentioned earlier. And watch plus/minus signs.

28kg*a=m*a+Fg ?

what's this??

You should be able to solve for Ft in the first equation for block m1 forces. Then solve for M2 using your equations for the block 2 forces.

 

[zombiengineer]

Ft acts up the plane, Fuk acts down the plane, and component of the weight acts down the plane, but it is not Fgcos53, you have a trig error. Also, you must ultimately calculate Fg and Fuk. Trig error in the weight component perpendicular to the plane. Yes, and you can determine 'a' from the kinematic equations. m is m2.
yes I think you mean Ft and not Fg after the equal sign. Also watch subscripts, it is m2. also, correct that Fgcos53 term as mentioned earlier. And watch plus/minus signs. what's this??

You should be able to solve for Ft in the first equation for block m1 forces. Then solve for M2 using your equations for the block 2 forces.

Thanks!

So after going down the line, and fixing that trig error, i got to this
Fn=ma*cos53.1
Fn=12kg*a

Fgsin53.1+(uk*12kg*a)=2*ma

(ma*.8)+(4.8kg*a)=2*ma
[cancel out the acceleration varible]

.8m+4.8kg=2m

4.8kg=1.2m
and
4.0kg=m

 

[PhanthomJay]

Thanks!

So after going down the line, and fixing that trig error, i got to this
Fn=ma*cos53.1

But Fn = Fgcos 53.1, where Fg does not equal ma. Fg is the weight,per your relevant equation.

Fn=12kg*a

The mass m1 is not 12 kg. m1 is 20 kg. Calculate its acceleration knowing that it moves 12 m in 3 s

Fgsin53.1+(uk*12kg*a)What happened to Ft?[/color] = 2*ma

You are getting careless, the mass is 20 kg . And 'a' is calculated from the kinematic equations.

(ma*.8)+(4.8kg*a)=2*ma
[cancel out the acceleration varible]

.8m+4.8kg=2m

4.8kg=1.2m
and
4.0kg=m

How can 4.0kg = 1 meter? You should realize that the units don't make sense, and that the equation is incorrect.

 

[zombiengineer]

But Fn = Fgcos 53.1, where Fg does not equal ma. Fg is the weight,per your relevant equation. The mass m1 is not 12 kg. m1 is 20 kg. Calculate its acceleration knowing that it moves 12 m in 3 s You are getting careless, the mass is 20 kg . And 'a' is calculated from the kinematic equations. How can 4.0kg = 1 meter? You should realize that the units don't make sense, and that the equation is incorrect.

Doh, I'm sorry, that was a horrible use of varibles on my part.

It should have been written as m= mass of block 2

the final answer reads mass of black 2 = 4.0 kg

as for the Fn=Fg cos 53.1 , another miswriting on my part,
as sumF in the y direction = Fn - Fg cos 53.1 = 0, therefore Fn= Fg cos 53.1
= mass of block 1 * acceleration * cos 53.1 = 12kg * acceleration

since Ft of block 1 = Ft of block 2
i can set the equations i have equal to each other and solve for mass of block 2, right?

getting this equation
Fgsin53.1+(uk*12kg*a)=2*ma
where m = mass of block 2, a= acceleration of the system, and the 12kg * a = the normal force, which is multiplied by the coefficent of kenetic friction, added to Fg sin 53.1 set equal to 2 * ma ( which is the tention of block two, which has no acceleration and a constant velocity)

after canceling out acceleration, i got
.8m+4.8kg=2m

where m=mass of block 2

then simply solved

Did I break physics there?
It looked to me like it would have worked.

 

[PhanthomJay]

Doh, I'm sorry, that was a horrible use of varibles on my part.

It should have been written as m= mass of block 2

the final answer reads mass of black 2 = 4.0 kg

as for the Fn=Fg cos 53.1 , another miswriting on my part,
as sumF in the y direction = Fn - Fg cos 53.1 = 0, therefore Fn= Fg cos 53.1
= mass of block 1 * acceleration * cos 53.1 = 12kg * acceleration

since Ft of block 1 = Ft of block 2
i can set the equations i have equal to each other and solve for mass of block 2, right?

getting this equation
Fgsin53.1+(uk*12kg*a)=2*ma
where m = mass of block 2, a= acceleration of the system, and the 12kg * a = the normal force, which is multiplied by the coefficent of kenetic friction, added to Fg sin 53.1 set equal to 2 * ma ( which is the tention of block two, which has no acceleration and a constant velocity)

after canceling out acceleration, i got
.8m+4.8kg=2m

where m=mass of block 2

then simply solved

Did I break physics there?
It looked to me like it would have worked.

You've got to proceed step by step, inch by inch...
1.) Calculate the acceleration of the hanging block knowing that the hanging block drops from rest a distance of 12 m in 3 s. Use one of the kinematic equations to solve for this downward acceleration. Write down your answer, a = ___?__.
2.) Since the cord forces the blocks to always move together, the down acceleration of the hanging block must be the same as the acceleration up the incline of the block on the incline, a = ___?__.
3.) Draw a free body diagram of the hanging block, identify the forces acting on it (Ft acts up and m₂g acts down) and write an equation using Newton's 2nd law
4.) Draw a free body diagram of the block on the incline, identify the friction, gravity, and tension forces acting on it along the incline, and write another equation using Newton 2 again. Solve for Ft.
5.) Plug Ft into your equation in (3.) above for the hanging block to solve for m₂.

Your present numbers and equations are all jumbled.