1. Today’s Concept: Friction
Physics 221 Lecture 6
Today’s Concept:
Friction

2. Stuff you asked about:
How is the angle between the horizontal and the ramp the same as the angle used in figuring out the components of gravity??? 90 q
90-q
mgcos(q) 90 q x y
mgsin(q) mg 09

3. Friction Direction: Always opposes the relative motion of two surfaces
Magnitude: Kinetic fk=mkN
Static: Does what needs to do up to a point:
Demos: static/kinetic friction
Incline plane/friction blocks 11

4. Checkpoint
A box sits on the horizontal bed of a moving truck. Static friction between the box and the truck keeps the box from sliding around as the truck drives. If the truck moves with constant acceleration to the left as shown, which of the following diagrams best describes the static frictional force acting on the box: S a A B C 13

5. Checkpoint S a
If the truck moves with constant acceleration to the left as shown, which of the following diagrams best describes the static frictional force acting on the box: A B C
A) In order to keep the box from sliding to the back of the truck as it accelerates, the frictional force needs to pull/push the box forward.
B) Friction always opposes motion/acceleration. 15

6. Checkpoint S a
If the truck moves with constant acceleration to the left as shown, which of the following diagrams best describes the static frictional force acting on the box: A B C
The force has to go in the same direction as the acceleration.
The static friction force is the NET FORCE on the box. 17

7. Checkpoint S a
The static friction force opposes the motion and so it acts in the opposite direction of the net force. No, the friction force IS the net force. The vertical forces add to zero.
The static friction force moves in the direction opposite to the direction of movement because it prevents the object from moving. Therefore, if the truck is moving to the left, the static friction force must be to the right. No, what would cause the block to accelerate along with the truck?
Forward because of the acceleration of the truck Yes.
If the bed of the truck is frictionless, the box moves to the right(in reference to the truck). To keep the box still, there must be a frictional force acting on it toward left. Very reasonable response. Ask, “what if ?”
I guessed. Is it any clearer now?
The box will move backwards so the frictional force must be opposite. But the box does NOT move backwards.
Should be in the same direction as the acceleration.
F = m a Net force and acceleration are always in the same direction. 17

8. ACT
A box of mass M sits on a horizontal table. A horizontal string having tension T applies a force on the box, but static friction between the box and the table keeps the box from moving. What is the magnitude of the net force acting on the box? M f T
A) Mg B) mMg C) T
D) 0
Since acceleration is zero. 19

9. Checkpoint
A box of mass M sits on a horizontal table. A horizontal string having tension T applies a force on the box, but static friction between the box and the table keeps the box from moving.
What is the magnitude of the static friction force acting on the box? M f T
A) Mg B) mMg C) T
D) 0
Using Newton's second law, we know that the force of tension+force of friction=0, therefore they are equal (and opposite). 20

10. Checkpoint
What is the magnitude of the static friction force acting on the box? M f T
Until the force exceeds the max static friction, the force of friction is equal and opposite to the horizontal force acting on it, resulting in a horizontal net force of 0.
The box is in equilibrium so the friction and tension forces must be equal and opposite.
T-f = 0 and so f = T since the box is not moving.
There are only two forces acting in the horizontal direction. Since there is no acceleration here, the sum of the forces must be zero so they must be equal but in opposite directions
Since the box is not moving, the static frictional force should equal to T.
I guessed. Do you have any more confidence or understanding now?
Since there is no acceleration the forces must be equal. 20

11. 22

12. Checkpoint m2 m2 Case 1 g Case 2 g m1 m1
Can we review systems that have pulleys, blocks and strings?
I can't quite remember what to do exactly.
A block slides on a table pulled by a string attached to a hanging weight. In Case 1 the block slides without friction and in Case 2 there is kinetic friction between the sliding block and the table. m2 m2
Case 1
(No Friction) g
Case 2
(With Friction) g m1 m1
In which case is the tension in the string biggest?
A) Case B) Case C) Same
75% got this right 23 12

13. m2 m2
Case 1
(No Friction) g
Case 2
(With Friction) g m1 m1
In which case is the tension in the string biggest? A) Case 1 B) Case 2 C) Same
B) M1 will not be accelerating as fast in case two as in case one because there is the extra force of friction acting against mass 1. Since the acceleration is smaller in case two, there has to be more of a force acting against gravity, the only other possible force is Tension. 26 13

14. In which case is the tension in the string biggest?m2 m2
Case 1
(No Friction) g
Case 2
(With Friction) g m1 m1
In which case is the tension in the string biggest?
A) Case B) Case C) Same
Some popular wrong explanations for picking B) :
Without friction there is no tension.
In case 1, m1 is like free fall, so there is no tension in the string. In case 2, the tension of the string is equal to the kinetic friction. 14

15. ACT
A block slides on a table pulled by a string attached to a hanging weight. In Case 1 the block slides without friction and in Case 2 there is kinetic friction between the sliding block and the table. m2 m2
Case 1
(No Friction) g
Case 2
(With Friction) g m1 m1
What is the tension in the string in Case 1?
A) T = B) T = m1g C) T is between 0 and m1g 25 15 15

16. ACT
A block (m2) slides on a table pulled by a string attached to a mass (m1) hanging over the side. The coefficient of kinetic friction between the sliding block and the table is mk. How are the tensions in the strings related? T2 m2 T1 g m1
What is the relationship between the magnitude of the tension of the string at block 2 and the magnitude of the tension in the string at block 1?
A) T1 > T2 B) T1 = T2 C) T1 < T2 16

17. ACT
A block (m2) slides on a table pulled by a string attached to a mass (m1) hanging over the side. The coefficient of kinetic friction between the sliding block and the table is mk. What is the tension in the string? a2 m2 g m1
What is the relationship between the magnitudes of the acceleration of the two blocks?
A) a1 = a2 B) a1 < a2 C) a1 > a2 a1 31 17

18. Lets work it out
A block (m2) slides on a table pulled by a string attached to a mass (m1) hanging over the side. The coefficient of kinetic friction between the sliding block and the table is mk. What is the tension in the string? a2 m2 g m1 a1
I'd like to have some more examples of problems with friction. 27 18

19. Lets work it out
A block (m2) slides on a table pulled by a string attached to a mass (m1) hanging over the side. The coefficient of kinetic friction between the sliding block and the table is mk. What is the tension in the string? a m2 g m1 a 27 19

20. a m2
1) FBD a m2 g m1 m1 a a 20

21. a m2
1) FBD N a T m2 g f T m1 m1 a
m2g a
m1g 21

22. a m2 1) FBD 2) SF=ma N a T m2 g f T m1 m1 a m2g a Block 2 m1g Block 1
N = m2g
T – m m2g = m2a
m1g – T = m1a
add
m1g – m m2g = m1a + m2a
m1g – m m2g
a =
m1 + m2 22

23. T is smaller when a is bigger
1) FBD
2) SF=ma N T m2 g f T m1 m1
m2g
m1g
m1g – T = m1a
m1g – m m2g
T = m1g – m1a
a =
m1 + m2
T is smaller when a is bigger 23