A problem involving Newton's second law.

[Clever_name]

Homework Statement

see this link http://i45.tinypic.com/10gwhw2.jpg

Homework Equations

F=kx, F=ma

The Attempt at a Solution

ok first I find the total length of the pulley system,
2Sa+Sb=L, differentiating with respect to time i get,
2Va + VB=0 and thus 2Aa+Ab =0
So Aa =-Ab/2

next I first analyze block A and do a force balance to get,
-Ma*g - k*x + 2T = Ma*Aa

following this i analyze Block B doing a force balance again,

T-Mb*g = Mb*Ab

so to eliminate one of the variables I input Aa=-Ab/2 into the equation describing block A.

I get -Ma*g-k*x+2T+Ma(-Ab/2)

then i get -2(Ma*g-k*X+2T)=Ab

so now i substitute this in Block B with real values

T-98.1=10(19.62-400+2T)
so T = 195.04
then Ab = 19.62-400+2(195.04) = 9.7 m/s^2

So now I have (delta)y = 0.5 m A=9.7
using the following equation i can solve for t, Δy=(1/2)at^(2)
doing so i get 0.32s
thus v = at = 9.7(0.32) =3.104 m/s

However i do not have a solution for this exercise and unsure if my attempt is correct or not. Any help is greatly appreciated thanks!

 

[gneill]

Since the length of the spring changes with time, the force will vary with time, too. That means the accelerations will not be constant, so you can't just apply the constant acceleration kinematic equations. Instead, you'll end up with a differential equation to solve.

Have you considered using a conservation of energy approach? Muuuuch simpler.

 

[Clever_name]

Ok so I assume my working is correct up to the point with which i calculated T.

so my new attempt is as follows:

W=(1/2)*m*vf^(2) - (1/2)*m*vi^(2)
F*D= (1/2)*m*vf^(2) - (1/2)*m*vi^(2)

so F*D = (-Ma*g-k*x+2T)*Δy = (1/2)*mvf^(2)

subbing in values and solving for vf i get 0.98m/s^2

I'm guessing however this is incorrect because the force that the spring exerts is not constant?

 

[gneill]

Ok so I assume my working is correct up to the point with which i calculated T.

so my new attempt is as follows:

W=(1/2)*m*vf^(2) - (1/2)*m*vi^(2)
F*D= (1/2)*m*vf^(2) - (1/2)*m*vi^(2)

so F*D = (-Ma*g-k*x+2T)*Δy = (1/2)*mvf^(2)

subbing in values and solving for vf i get 0.98m/s^2

I'm guessing however this is incorrect because the force that the spring exerts is not constant?

Yup. This approach must yield a differential equation. The spring force is not constant, so the accelerations are not constant and you cannot apply the constant acceleration kinematic formulae.

Try an energy conservation approach.

 

[Clever_name]

Ok my next attempt is as follows

(1/2)ky^(2) = Mb*vf^(2)

so subbing in values to solve for vf i get 4.472m/s.

 

[Clever_name]

actually maybe this is the equation i need?

(1/2)ky^(2) +Ma*g*y=(1/2)Mb*Vf^(2)
and solving i get 4.686 m/s

 

[gneill]

Suppose you set the individual gravitational PE's of the two masses to zero when the system is at equilibrium. Also set the spring's PE to zero at equilibrium. (You can do this because PE is a relative measure).

Now, whenever the system is at the equilibrium position, those PE's will be zero.

How much additional energy will be stored in the spring when mass B is displaced by 0.5m and held there? What's the total KE at that time? (Remember, the masses are stationary). So what's the total energy PE+KE at that time?

 

[Clever_name]

PEa = Ma*Aa*y and KEa = 0
PEb= Mb*Ab*y and KEb =0

I can't use g as acceleration right? because the forces acting on the objects interfere with its normal functioning? or am i way off the mark :(

 

[gneill]

PEa = Ma*Aa*y and KEa = 0
PEb= Mb*Ab*y and KEb =0

Okay, not what I asked. What's the change in PE stored in the spring when mass MB is displaced by 0.5m?

I can't use g as acceleration right? because the forces acting on the objects interfere with its normal functioning? or am i way off the mark :(

You won't need ANY accelerations or forces using the conservation of energy method.

 

[Clever_name]

ok so that would be -kx

 

[gneill]

ok so that would be -kx

No, What's the formula for the energy stored in a spring?

 

[Clever_name]

sorry, (1/2)ky^(2)

 

[gneill]

So... What's the additional energy stored in the spring when mass B is displaced?

 

[Clever_name]

(1/2)Mbvf^(1/2)

 

[gneill]

No. There's no motion at this point. Mass B is displaced and held stationary at its new position. But the spring has been stretched. What energy does SPRING gain?

 

[Clever_name]

potential energy

 

[gneill]

Make an effort. Calculate the value!

 

[Clever_name]

Ok I'm thinking, (1/2)k(xf^(2)+xi^(2))=1/2(Mb)vf^(2)

subbing in values i get

6.32 m/s?

 

[gneill]

Ok I'm thinking, (1/2)k(xf^(2)+xi^(2))=1/2(Mb)vf^(2)

subbing in values i get

6.32 m/s?

How do you justify the equation above? (What does it represent?)

What about mass A? Won't it be moving too?

The reason I ask you to calculate the potential energy added to the spring when the displacement is carried out is that it represents the energy added to the overall system, and will stay with the system when it returns to the equilibrium position. When the system returns to the equilibrium position all the gravitational potential energy changes due to displacement disappear -- everything is back in its original position. What's left is the new energy that was stored in the spring, which has then been released as KE since the spring, too, is back in its original position.

So why not calculate that spring energy? How much is the spring extended when MB is displaced by 0.5m? How much energy does that put into the spring, in Joules?

 

[Clever_name]

" How much is the spring extended when MB is displaced by 0.5m?"
the spring is extended 0.5m.

"How much energy does that put into the spring, in Joules?"
(1/2)k(y)^(2) = (1/2)*800*(0.5)^2 = 100 J

 

[Clever_name]

ok new attempt total potential energy = (1/2)kx^(2)+(1/2)*kx^(2) = (1/2)MbVb^(2) -(1/2)MaVa^(2)

we know Va = -Vb/2
so
(1/2)kx^(2)+(1/2)*kx^(2) = (1/2)MbVb^(2) -(1/4)MaVb^(2)

then we get 2kx^(2)=(Mb-(1/2)Ma)Vb^(2)
and i need to solve this equation?

 

[Clever_name]

my reasoning is as follows:
block A
(1/2)kx^(2)-(1/2)MaVa^(2)= 0
and
block b
-(1/2)(kx^2)+(1/2)MbVb^(2)=0

 

[gneill]

Confirm how far the spring is stretched when the mass Mb moves by a given amount. The pulley arrangement provides a mechanical advantage...

 

[Clever_name]

ok so 2SA+SB=L

then SB=-2SA

so SB/-2 = SA then SA=-0.25m=xa
so
(1/2)kxa^(2)+(1/2)*kxb^(2) = (1/2)MbVb^(2) -(1/4)MaVb^(2)
subbing in values and simpilfying i get 4.5Vb^(2)-125=0
solving i get 5.27 m/s

 

[Clever_name]

actually tho won't the spring get stretched by the same amount as the displacement at B since the cord is inextensible?

 

[Clever_name]

actually disregard post 24 its nonsense.

 

[Clever_name]

ok so (delta)L = 2SA +SB

(delata)L = 0.5m

so Sb=0.5-2(0.5) = -0.5m = xa

subbing this into (1/2)kxa^(2)+(1/2)*kxb^(2) = (1/2)MbVb^(2) -(1/4)MaVb^(2)
and solving for vb i get 6.66m/s

 

[gneill]

Nearly there.

Keep in mind that kinetic energy doesn't have a direction. All kinetic energies for a system are positive and sum to the total KE.

For the PE added to the spring when the displacement occurs, take the difference between the PE stored in the spring initially and the PE at its full extension. That means finding the initial (equilibrium) stretch (x_i) and the final stretch (x_f) and taking the difference in the PE's at those positions of the spring:
$$\Delta PE = \frac{1}{2} k x_f^2 - \frac{1}{2} k x_i^2$$
The initial (equilibrium) stretch ##x_i## can be found by balancing the forces at the top of the spring when the system is in equilibrium -- the spring's force and the weight of Ma balances the upward force applied by Mb through the pulley system.

 

[gneill]

Be sure to take a close look at the dynamics of the pulley system motions. If there's a mechanical advantage involved, there will always be a similar ratio of motions involved. You've already identified that the force applied to mass A through string tension is doubled due to the two string segments holding up the Mass A pulley.

Mass A moves half the distance that mass B moves. Similarly, mass A's velocity and acceleration rates are half that of mass B.

 

[Clever_name]

ok so my attempt at finding xf and xi is as follows:
finding xi,
kxi-Ma*g+2T = 0

we know T = Mb*g =98.1
so subbing into equation and solving i get -0.05

now finding xf
-kxf-Ma*g+2T = 0 at the instant the string as gone 0.5m.
i solved for T previously and now can directly solve for xf
doing so i get 0.51m

the (delta)P = (1/2)(800)(0.22)^(2)-(1/2)(800)(0.05)^(2) = 18.36

 

[gneill]

Those values don't look right.

When the system is static at equilibrium, mass Mb pulls on the string with weight Mb*g. So that's the tension in the string, as you've written.

Twice this force (2T) is balanced by the weight of Ma and the force due to the spring. So write

2T = Ma*g + k*xi

Find xi.

You don't need any force or tension equations to find the amount by which the spring length changes when Mb is displaced. The displacement of Mb is coupled to the displacement of Ma and the spring via the pulley system. How much does Ma move for a given movement of Mb?

 

[Clever_name]

2T=Ma*g +k*xi,
subbing in values xi = 0.27m

now this is the part where I'm confused- if the string is inextensible then isn't the displacement of Ma and the spring the same value as the displacement of Mb?

 

[gneill]

2T=Ma*g +k*xi,
subbing in values xi = 0.27m <--- Looks a bit high. Check the calculation.

now this is the part where I'm confused- if the string is inextensible then isn't the displacement of Ma and the spring the same value as the displacement of Mb?

No. There is a mechanical advantage involved, since there's two segments holding up the Ma pulley and only one holding the Mb mass. Check the operation closely. Draw a detailed figure with accurate lengths and measure them if required (choose your own scale and initial lengths).

 

[Clever_name]

2SA=SB, that is to say that the vertical motion of B is twice the vertical motion of A

for the calculation i got

(2(98.1)-2(-9.81))/800 = 0.27m

 

[gneill]

2SA=SB, that is to say that the vertical motion of B is twice the vertical motion of A

for the calculation i got

(2(98.1)-2(-9.81))/800 = 0.27m

Why the "-" on the 9.81? At the top of the spring, the 2T from Mb is pulling upwards, while Ma and the spring pull downwards. Considering force magnitudes, Ma*g should sum with k*xi.