1. ## help!

If the coefficient of kinetic friction between the block and the table in the figure below is 0.560, and m1 = 0.150 kg and m2 = 0.250 kg (a) what should m3 be if the system is to move with a constant speed? (b) If m3 = 0.100 kg, what is the magnitude of the acceleration of the system? (Assume ideal conditions for the string and pulleys.)

(m3)
|----------|
( m1) | | (m2)

2. Originally Posted by Refujoi
If the coefficient of kinetic friction between the block and the table in the figure below is 0.560, and m1 = 0.150 kg and m2 = 0.250 kg (a) what should m3 be if the system is to move with a constant speed? (b) If m3 = 0.100 kg, what is the magnitude of the acceleration of the system? (Assume ideal conditions for the string and pulleys.)

(m3)
|----------|
( m1) | | (m2)
I can't seem to figure out where the pulley is in the problem. Could you please describe the physical situation in words?

-Dan

3. ## Help!

Originally Posted by Refujoi
If the coefficient of kinetic friction between the block and the table in the figure below is 0.560, and m1 = 0.150 kg and m2 = 0.250 kg (a) what should m3 be if the system is to move with a constant speed? (b) If m3 = 0.100 kg, what is the magnitude of the acceleration of the system? (Assume ideal conditions for the string and pulleys.)

(m3)
|-----------------------|
( m1) (m2)

I can't seem to figure out where the pulley is in the problem. Could you please describe the physical situation in words?

-Dan

There is a table left side is (m1) top (m3) right (m2) m1 and m2 are get the floor

4. If the coefficient of kinetic friction between the block and the table in the figure below is 0.560, and m1 = 0.150 kg and m2 = 0.250 kg (a) what should m3 be if the system is to move with a constant speed? (b) If m3 = 0.100 kg, what is the magnitude of the acceleration of the system? (Assume ideal conditions for the string and pulleys.)

(m3)
( m1)-----------------------(m2)

5. Originally Posted by Refujoi
Originally Posted by Refujoi
If the coefficient of kinetic friction between the block and the table in the figure below is 0.560, and m1 = 0.150 kg and m2 = 0.250 kg (a) what should m3 be if the system is to move with a constant speed? (b) If m3 = 0.100 kg, what is the magnitude of the acceleration of the system? (Assume ideal conditions for the string and pulleys.)

(m3)
|-----------------------|
( m1) (m2)

I can't seem to figure out where the pulley is in the problem. Could you please describe the physical situation in words?

-Dan

There is a table left side is (m1) top (m3) right (m2) m1 and m2 are get the floor
Gotcha!

Okay. There's a bit of a shortcut for this, but it's probably better to see it the long way.

Part a)
Three free-body diagrams are required.
m1: Typically I choose the direction of acceleration to be positive, but since a = 0 there isn't one. So I will arbitrarily choose +y upward here. We have a tension acting upward, call it T1. There is a weight acting downward, call it w1 = (m1)g. Newton's 2nd tells us:
$\displaystyle \sum F_1 = T_1 - m_1g = m_1a = 0$ since a = 0. So $\displaystyle T_1 = m_1g$

m2: There's still no acceleration but we want to keep the directions consistent. Thus if m1 is "accelerating" upward in FBD#1 we should choose +y to be down to be consistent. So we've got a tension T2 pointing up and a weight w2 = (m2)g pointing down.
$\displaystyle \sum F_2 = w_2 - T_2 = m_2a = 0$ (Note the + direction!) So $\displaystyle T_2 = m_2g$

m3: For the same consistency reason as in FBD#2 we want +x to the right. We need an upward direction here as well. I'll choose +y upward. So we've got a normal force N3 acting upward, a weight w3 = (m3)g acting downward, a tension force T1 acting to the left (by Newton's 3rd this must be equal to the tension force acting on m1), a tension force T2 acting to the right (ditto for T2), and a kinetic friction force fk acting...which direction should it act? We've got that there's no acceleration, but the system is moving at constant v. The mass of m3 should depend on this direction! I'm going to set up TWO Newton's 2nd equations for this, the first with the kinetic friction acting to the right, then the second to the left.
(1)$\displaystyle \sum F_{3x} = T_2 - T_1 +f_k = m_3a = 0$

(2)$\displaystyle \sum F_{3x} = T_2 - T_1 -f_k = m_3a = 0$

$\displaystyle f_k = T_1 - T_2 = m_1g - m_2g = (m_1 - m_2)g$
Now, m2 = 0.250 kg and m1 = 0.150 kg. This means that fk is negative. But the way we've set up the Newton's 2nd forces all the variables are positive. This means that equation 1 cannot happen! Thus the system of blocks can only be moving to the left with constant speed.

We also get another Newton's 2nd from FBD#3:
$\displaystyle \sum F_{3y} = N_3 - m_3g = 0$ since block 3 is obviously not accelerating upward.

To sum up:
a. $\displaystyle T_1 = m_1g$
b. $\displaystyle T_2 = m_2g$
c.$\displaystyle f_k = T_2 - T_1 = m_2g - m_1g = (m_2 - m_1)g$ (from equation (2) above)
d.$\displaystyle N_3 = m_3g$

and don't forget: e. $\displaystyle f_k = \mu _k N_3$

Take equation c and do some substituting:
$\displaystyle \mu _k N_3 = (m_2 - m_1)g$ (from equation e.)

$\displaystyle \mu _k (m_3g) = (m_2 - m_1)g$ (from equation d.).

Thus
$\displaystyle m_3 = \frac{(m_2 - m_1)g}{\mu _k g} = \frac{(m_2 - m_1)}{\mu _k}$

Part b)
This is practically identical to the above. I'll leave it to you. Keep track of the (now) non-zero acceleration. You'll have to guess which direction the system will accelerate in, but it shouldn't be too hard to figure out. (And, of course, all the blocks will have the same acceleration since both strings need to stay under tension.) Give it a try and if you can't figure it out, post here and I'll take you through it.

-Dan