Transcript Tutorial_blocks

Two blocks (m1 = 5kg, m2 = 2.5kg) are in contact
on a frictionless table. A constant horizontal force
FA = 3N is applied to the larger block as shown.
Find the magnitude of the force F1 on 2 exerted by
the larger block on the smaller block.
Find the magnitude of the force F2 on 1 exerted by
the smaller block on the larger block.
m1
FA
m2
y
x
•
•
•
•
•
•
•
•
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Question 8
Qualitative analysis
1. Which of the following statements
is true?
• A : This is an equilibrium problem, the two blocks remain
at rest.
• B : This is an equilibrium problem, the two blocks move
at a constant velocity.
• C : This is a non-equilibrium problem, the two blocks
move at a constant velocity.
• D : This is a non-equilibrium problem, the two blocks
accelerate.
Choice: A
Incorrect
Since our blocks are on a frictionless table, there is no
friction force resisting the force that is applied to the larger block.
Only if there was a friction force equal to the applied force would
the system by in equilibrium.
-Equilibrium requires that no net force acts on object
- Otherwise, the object is not in equilibrium.
- If velocity is constant (or zero) then object is in equilibrium.
Choice: B
Incorrect
Since our blocks are on a frictionless table, there is no
friction force resisting the force that is applied to the larger block.
Only if there was a friction force equal to the applied force would
the system by in equilibrium.
-Equilibrium requires that no net force acts on object
- Otherwise, the object is not in equilibrium.
- If velocity is constant (or zero) then object is in equilibrium.
Choice: C
Incorrect
Since our blocks are on a frictionless table, there is no
friction force resisting the force that is applied to the larger block.
Only if there was a friction force equal to the applied force would
the system by in equilibrium.
-Equilibrium requires that no net force acts on object
- Otherwise, the object is not in equilibrium.
- If velocity is constant (or zero) then object is in equilibrium.
Choice: D
Correct
This is a true statement. Since our blocks are on a
frictionless table, there is no friction force resisting the force that is
applied to the larger block. Only if there was a friction force equal to
the applied force would the system by in equilibrium.
The blocks will accelerate, and our system is not in equilibrium.
Qualitative analysis
2. Which of the following statements is true?
• A : The applied force will act on both blocks because the
smaller block is in the way of the larger block.
• B : The applied force will act only on the larger block.
The smaller block will feel a smaller horizontal force
exerted by the larger block.
• C : The acceleration of the two blocks will be different
because they have different masses.
Choice: A
Incorrect
The applied force to our blocks, perhaps by
someone’s hand, is a type of contact force.
A contact force acts only at the place of contact (it
does not act on objects it is not touching).
Choice: B
Correct
This is true because the force applied to our blocks,
perhaps by someone’s hand, is a type of contact force.
A contact force acts only at the place of contact (it does
not act on objects it is not touching).
Choice: C
Incorrect
If objects move together, they will
have the same acceleration.
a1
a2
a
a  a1  a2
3. Newton’s 3rd Law, in general, states:
“If body A exerts a force on body B (action), then body B exerts a force
on body A (reaction) that is equal in magnitude and opposite in
direction.”
Considering this law, what do we know is true about our
situation?
• A:
• B:
• C:
• D:
There is a force exerted on the larger block by the smaller
block (F2 on 1), and there is also a force exerted on the
smaller block by the larger block(F1 on 2).
The direction of F2on1 should be opposite of F1 on 2.
|F2 on 1| = |F1 on 2|
All of the above are true.
Choice: A
This is not the only correct choice.
This is true according to Newton’s third law.
Since block 1 is pushed into the block 2 there is
a force (action) on block 1 by block 2. Block two
must exert a force (reaction) on block 1.
Choice: B
This is not the only correct choice.
This is true according to Newton’s third law.
Since block 1 exerts a force (action) on block 2
by being pushed into it, block two must exert a
force (reaction) on block 1.
This pair of forces, as with all action-reaction pairs,
act in opposite directions, and have the same
magnitude.
Choice: C
This is not the only correct choice.
This is true according to Newton’s third law. Since
block 1 exerts a force (action) on block 2 by
being pushed into it, block two must exert a
force (reaction) on block 1.
These two forces, as with all action-reaction pairs,
have equal magnitudes.
Choice: D
Correct
Since block 1 exerts a force (action) on block
2 by being pushed into it, block two must exert a
force (reaction) on block 1.
This pair of forces, as with all action-reaction pairs,
act in opposite directions, and have the same
magnitude.
Planning
4. Which pair of free-body diagrams are correct ?
• A:
N1
F2on1
• B:
N2
FA
m2 g
N1
N2
FA
m1 g
• D:
F1 on 2
m1 g
F2on1
• C:
F2 on 1
N1
F2on1
FA
Bold type
denotes vector
quantities in
all free-body
diagrams.
FA
m2 g
N2
FA F1 on 2
m1 g
m2 g
N1
N2
F2on1
FA
m1 g
F1 on 2
m2 g
Choice: A
Incorrect
There is no force that pushes the
smaller red block from the left.
Choice: B
Incorrect
The applied force does not act on the 2nd
smaller red block, but there is a force from the
1st block pushing on the 2nd block.
Choice: C
Incorrect
The applied force does not act on the 2nd smaller block.
Choice: D
Correct
These diagrams represent all of the forces on each
block correctly.
Note that we can also make a free-body diagram
for the two blocks together as a system:
N
FA
(m1+m2)g
Planning
5. Which one of the following statements is
incorrect about the vertical motion or vertical
forces acting on the blocks?
A: None of the vertical forces are relevant for solving this problem.
B: Vertical and horizontal motions are completely decoupled in this
problem since the surface is frictionless.
C: The solution to this problem will be the same on the earth and on the
moon because the gravitational force is not important for this problem.
D: Gravitational force is important for solving the problem.
Choice: A
Incorrect
This statement is correct.
We are only concerned with the forces and
motion in the positive and negative x-directions
in this problem, because the vertical forces
cancel out. The force of gravity on the blocks is
canceled by the normal force.
Choice: B
Incorrect
This statement is correct.
We can solve this problem by only
analyzing the forces in the x-directions.
Choice: C
Incorrect
This statement is correct.
We can solve this problem by only
analyzing the forces in the x-directions.
The force of gravity act in the negative ydirection and is cancelled by the normal
force.
Choice: D
Correct
This statement is incorrect, because gravity
pulls in the negative y-direction and is exactly
canceled by the normal force acting in the
positive y-directon.
Gravitational force is not important for solving
the problem.
Implementation
6. Apply Newton’s 2nd Law to each block individually, and
then to the two blocks together as a system.
Which of the following equations is not a result of applying
Newton’s 2nd Law? (a is acceleration)
• A: FA – F 2 on 1 = m1a
• B: FA – F1 on 2 = (m1 + m2)a
• C: F1 on 2 = m2a
• D : FA = (m1+m2)a
Choice: A
Incorrect
We get this equation be applying Newton’s
2nd Law to the larger block:
Recall the free body diagram for the first block from question 3:
N1
F2 on 1
FA
m1 g
Use Newton’s 2nd Law to sum up the forces in each direction:
Fx  m1ax
FA  F2 on 1  m1ax
Fy  m1ay  0
m1g  N1  0
No motion in
the y-direction

Choice: B
Correct
Applying Newton’s 2nd Law to either block or both
together will not lead to this equation.
Choice: C
Incorrect
We get this equation be applying Newton’s
2nd Law to the smaller block:
Recall the free body diagram for the second block from question 3:
N2
F1 on 2
m2 g
Use Newton’s 2nd Law to sum up the forces in each direction:
Fx  m2ax
F1 on 2  m2ax
Fy  m2ay  0
m2g  N2  0
No motion in
the y-direction

Choice: D
Incorrect
1.
We get this equation be applying Newton’s
2nd Law to the system of blocks moving
together:
Make a free-body diagram for the 2 blocks together:
N
FA
(m1+m2)g
2.
Use Newton’s 2nd Law to sum up the forces in each direction:
Fx  ma x
Fy  ma y  0
m  m1  m2
m1  m2 g  N  0
FA  m1  m2 ax

No motion in
the y-direction
7. Now use the correct equations that we
found in the previous question to solve for
the magnitude of the force (F1 on 2) exerted
on the smaller block by the larger block.
Answer
1st: Solve for acceleration
2nd: Substitute to find F1 on 2
FA  m1  m2 ax
FA
 ax
m1  m2 
F1 on 2  m2ax
F1 on 2  2.5kg0.4m/s2 
F1 on 2  1N

3N
 ax
5kg  2.5kg
0.4m/s2  a x

8. Recall from using Newton’s 3rd Law in
question 3 that |F2 on 1| = |F1 on 2|, so the magnitude of
F2 on 1 is also 1 N.
Let’s check that this is true using the correct
equations that we found in the previous
questions to solve for the magnitude of the
normal force (F2 on 1) exerted on the larger block
by the smaller block.
Answer
ax  0.4m/s2

FA  F2
on 1
F2
 m1ax  FA
on 1
 m1ax
F2
on 1
 5kg0.4m/s  3N
F2
on 1
 1N
F2
on 1
 1N
2
Assessment
•Do units look correct?
•Is |F2 on 1| = |F1 on 2|?
•Is the direction of F2 on 1 opposite FA?
Reflection
•Newton’s 3rd law gave an important relation for the forces
between blocks. This law explains the normal forces that the
blocks exert on each other.
•Blocks moving together have the same acceleration.
•The applied force (FA) only acts on larger block.
•If the force FA were applied to the smaller block in the
opposite direction, the solution can be obtained by swapping
m2 and m1:
FA
FA
F1 on 2  m1
F2
on 1  m2
m1  m2

1
m2
m1
FA
FA

m2  m1 1 m1
m2