Mechanical Problem: Calculating Car Velocity at Time 0 & 5 m/s

[Fabio010]

Homework Statement

http://www.freeimagehosting.net/t/7bf5e.jpg

Ok the initial velocity of the car is 5 m/s to the left. Calculate for each situation, the instant time when the velocity of car is :

a)null and 5m/s to the right.



For situation one with just one pulley:

In the block 2 we have F = P-T (=) P = T+m(A)*a (1)
In car, because it is going to the right F = T (=) T = m(B)*a (2)

because he tension that block do in car is equal to the tension that car do in block so

F = P-T (=) F = P - (m(B)*a) (=) P = m(A)*a + m(B)*a (=) P = a(mA+mB)
a = (10*9.8)/(50+10) (=) a = 1.633 m/s^2


So with the law of velocity v = vi + at

v = 0 (null)
vi = 5m/s

a = - 1.633m/s^2 (negative because it is opposite to the car)

t(when v=0) = -5/-1.633 = 3.06s
t(when v = -5) = -10/-1.633 = 6.12s


For the situation with two pulleys


I do not know how to do it, because in the system of the two pulleys we have two tension and i know that they are T/2 each one

but my teacher said that the acceleration of the car is 2* acceleration of the box.

My problem is Why? and even knowing that a(A) = 2*a(B), how can i find the acceleration to the block 3 ?

I appreciate the help.

 

[Fabio010]

Sorry for double post. But please help.
I really need to do this problem :/

 

[tiny-tim]

Hi Fabio010!

…but my teacher said that the acceleration of the car is 2* acceleration of the box.

This isn't physics, it's just geometry …

the string has fixed length,

so if you make the distance between the pulleys increase by y, you use 2y of string, don't you?

so when the block moves down y, the car moves along x = 2y

differentiate once, or twice, and you get dx/dt = 2dy/dt, d²x/dt² = 2d²y/dt² ​

… even knowing that a(A) = 2*a(B), how can i find the acceleration to the block 3 ?

when you do F₁ = m₁a₁ and F₂ = m₂a₂,

you relate them by a₁ = 2a₂ ​

 

[Fabio010]

hum..

so i guess that for block 3 we have m3a3 = mg-T;
and for car we have T = m2*a2

(a2) = 2(a3)

But solving the equations to find the acceleration, it gives a wrong a solution.

 

[tiny-tim]

for block 3 we have m3a3 = mg-T;
and for car we have T = m2*a2

(a2) = 2(a3)

(try using the X² and X₂ buttons just above the Reply box )

that should work

show us your full calculations ​

 

[Fabio010]

"Damn sorry i just checked you tip after post"

ok

m₍₃₎a₍₃₎ = mg -Tas we know T = m₍₁₎*a₍₁₎
so:
m₍₃₎a₍₃₎ = m₍₃₎g - m₍₁₎*a₍₁₎

because a₍₁₎ = 2a₍₃₎

m₍₃₎a₍₃₎ = m₍₃₎g - m(2)*2a₍₃₎
a₍₃₎ (m₍₃₎+2m₍₁₎ ) = m₍₃₎g
a₍₃₎ = m₍₃₎g/((m₍₃₎+2m₍₁₎ ))

a₍₃₎ = 20*9.8/((20+100))
a₍₃₎ = 1.633 m/s^2

so a₍₁₎ = 3.2667 m/s^2

law of velocity of car is v = 5 -(3.2667)t
when v = 0 t = 1.55 s
solution t = 2.8 s

 

[tiny-tim]

oops!

so i guess that for block 3 we have m3a3 = mg-T

that should work

oops! i forgot to check the diagram!

it isn't mg - T, is it? (draw a free body diagram)

 

[Fabio010]

It is mg -2T because of the two pulleys :O

Now it is correct :)

Thanks for the help :)