What Is the Time for Log A to Move 1 Meter on a Diagonal Plane?

[JAZZ541]

Homework Statement

Log A weighs 20Kg is laid on a diagonal smooth plane with a diagonal angle of 27 37 degree, log A is tied by a rope to the head of the plane and there through a wheel with no friction the rope descend vertically to the bottom of the plane and through another friction-less wheel is tied to log B that weighs 20 Kg and is laid on a smooth horizontal plane' the last part of the rope tied to log B is horizontal

Homework Equations

1. what is the time in which log A will move a 1 meter distance on the plane if it starts from rest?
2. what is the tension in the rope attached to both logs?

The Attempt at a Solution

is acceleration here 0? or maybe 6? I'm a bit lost would appreciate any help, thx!

[edit: fixed typo in given angle (picked up correction from post #3 -- Moderator]

 

[Doc Al]

is acceleration here 0? or maybe 6?

Well, you know the acceleration can't be zero, else the first question would not make sense.

Do this: Analyze the forces acting on A and on B. Apply Newton's 2nd law to each.

 

[JAZZ541]

Ty!
Is this correct? btw angle is 37 and not 27

 

[Doc Al]

Well, you don't show arrows. (Forces have direction.) And what does "f" represent? How do t1 and t2 relate?

 

[JAZZ541]

f - friction ohh, but they said smooth surface, so no friction
and T1 = T2?

 

[Doc Al]

f - friction ohh, but they said smooth surface, so no friction

Right.

and T1 = T2?

Right. The tension is the same throughout the rope.

 

[JAZZ541]

I'm sorry I had to go ill try and work on it some more and come back later

 

[Doc Al]

I'm sorry I had to go ill try and work on it some more and come back later

No worries. Give it a shot and then come back and show what you've done.

 

[JAZZ541]

Thank u for your patience!
I have this:

1.

m1(A)g = m2(B)g = 20Kg

X = 1m

mag • sinα – T = m1a

T = m2a

mag • sinα -T + T = m1a + m2b

200 • 0.60 = a(m1 + m2)

a = 120/40 = 3m/s

X = V0T + at^2 /2

1 = 0 + 3t^2/2

2 = 3t^2

t = 0.816s

2.

T = m2a

T = 20 • 3 = 60N

there are two more questions about this problem:

3.
If there is a friction between log B and the horizontal surface what is the minimal static coefficient of friction in which if the system is in a state of rest then it will remain in that state?

4.
If the kinetic and static coefficient of friction is mu = 0.25, how long will it take log A to travel a 2 meter distance on the plane (the rope is weightless)?

how should I approach that?

Thx again!

 

[Doc Al]

m1(A)g = m2(B)g = 20Kg

X = 1m

mag • sinα – T = m1a

T = m2a

mag • sinα -T + T = m1a + m2b

200 • 0.60 = a(m1 + m2)

a = 120/40 = 3m/s

Good!
(1) Your first equation should be: $m_1 = m_2 = 20$ Kg
(2) Careful with units. Acceleration has units of m/s^2.
(3) It's almost always better to stick to symbols as long as possible before plugging in numbers. Here's how I'd do it:
Since the masses are the same, I'd use "m" for both masses. And since the acceleration is the same, I'd use "a" for that.
$mg\sin\theta - T = ma$
$T = ma$
Adding them, we get:
$mg\sin\theta = 2ma$
And so on.

Note that the mass cancels out! So it doesn't matter (at least for this part of the question).

X = V0T + at^2 /2

1 = 0 + 3t^2/2

2 = 3t^2

t = 0.816s

Good!

2.

T = m2a

T = 20 • 3 = 60N

Good!

 

[Doc Al]

3.
If there is a friction between log B and the horizontal surface what is the minimal static coefficient of friction in which if the system is in a state of rest then it will remain in that state?

(1) What additional force acts on B? (Use symbols, not numbers.)
(2) What is the acceleration?

Redo your force equations with this in mind.

4.
If the kinetic and static coefficient of friction is mu = 0.25, how long will it take log A to travel a 2 meter distance on the plane (the rope is weightless)?

This is similar to the first part, only now there is an additional force acting on B.

 

[JAZZ541]

TY!

(1) What additional force acts on B? (Use symbols, not numbers.)
(2) What is the acceleration?

a = 0 ?

Redo your force equations with this in mind.

mgsinα = T = f = mumg ?

 

[Doc Al]

Good!

(I fixed your quotes.)

 

[JAZZ541]

Then:
mu = sin37° = 0.6
Ty very very much for your patience and guidance!

 

[Doc Al]

Then:
mu = sin37° = 0.6

Yep!

Ty very very much for your patience and guidance!

You are welcome.