1. Consider the two masses hung over a frictionless pulley wheel
5 kg
2 kg

2. Consider the two masses hung over a frictionless pulley wheel
5 kg
2 kg

3. Consider the two masses hung over a frictionless pulley wheel
What is the acceleration of the masses?
2 kg
5 kg

4. What Equation Can We Use?
 F = ma
Solve for a:
a =  F/m
But What forces are acting on the masses? We don’t know the  F yet...
F2 = m2g
F1 = m1g

5. Net Force  F = F1 + F2  F = -29.4 N So  F = -m1g + m2g
Let’s call the direction of F1 negative and F2 positive…
So  F = -m1g + m2g
 F = -(5 kg)(9.8m/s/s)+(2kg)(9.8m/s/s)
 F = -49 N N
 F = N
= ma
F1 = m1g
F2 = m2g

6. So, ma = -29.4 N Recall, we are trying to figure out the acceleration.
So solving for a, we get a = (-29.4 N)/m
…But which m do we divide by?
The one that is accelerating. Which one is accelerating then?
BOTH OF THEM!!
…Therefore
F1 = m1g
F2 = m2g

7. Divide by the COMBINED mass
So once again, we get a = (-29.4 N)/m
a = (-29.4 N)/(m1+m2)
a = (-29.4 N)/(5 kg+2 kg)
a = (-29.4 N)/(7 kg)
a = 4.2 m/s/s
F1 = m1g
F2 = m2g

8. Now…How could you find the TENSION in the string?
IS there any tension in the string?
Which string has MORE tension? The right side of the pulley or the left of the pulley?
Is there any difference?
F1 = m1g
F2 = m2g

9. Now…How could you find the TENSION in the string?
The string on the right side of the pulley is the same string that is on the left side of the pulley…
That one string is attached to two masses that are both affected by gravity.
Even though they are not equal masses, isn’t there some degree of “tightness” in the string?
There is the same amount of tightness (tension) on both sides of the pulley
F1 = m1g
F2 = m2g

10. So how can you find that Tension?
Let’s look at one of the two masses for a moment...
F1 = m1g
F2 = m2g

11. Diagram all of the Forces
1st let’s show the gravitational Force because gravity always works in the vertical direction
Tension in the string is acting upward (keeping the block from going into “free fall”)
How about the direction of the Net Force? What was the direction of the acceleration? (from part 1 of this problem)
Tension a
Fg = m1g

12. Which Forces are Known? -  F = -Fg + T So Fg -  F = T
Weight (Fg) is known...
49N (negative direction)
Net force (ma) is known…
ma = 5 kg x 4.2 m/s/s =
= 21 N (negative direction)
Tension is the only unknown force in our diagram…
But  F = Fg + Tension
-  F = -Fg + T
So Fg -  F = T
49 N - 21 N = T
28 N = T
Fg = m1g
Tension
 F

13. Please do #92 on page 115