PHYS 110 Spring 2011
HW 6.6 SOLUTIONS
For this problem, consider the figure below. Two blocks, m1 and m2, are attached by a massless rope
which is draped over an ideal pulley. Block 2 is atop an inclined surface with an angle of inclination β.
Consider the following 5 cases:
m1
m2
β
Case I. – Frictionless surface, zero acceleration.
1. Draw a FBD for each block. Include a coordinate system for each FBD.
To start the free body diagrams, we can think about the following:

A dotted line drawn around the border of our system in the rough sketch can help
us find all the contact forces. What objects outside the system are in contact with
P? What type(s) of force are associated with this type of interaction?

Take into account any non-contact forces that cannot be easily determined by the
dotted line method.
Here is what we find for block 1:

Block 1 is only in contact with rope and the incline. The rope is exerting a tension
force in the direction away from the block towards the pulley. The incline is
exerting two forces on the person – a Normal force N1I (on block 1 by incline) and a
static frictional force f1I (on block 1 by the incline). These are the perpendicular
and parallel components of the interaction between the block and the incline.
WAIT! There is no friction for case I! For later cases, we will need to include that
though.

The only other force acting on the block 1 is the weight force W1E (on block 2 by
the Earth).
PHYS 110 Spring 2011
HW 6.6 SOLUTIONS
Here is what we find for block 2:

Block 2 is only in contact with rope. The rope is exerting a tension force in the
direction away from the block towards the pulley.

The only other force acting on the block 2 is the weight force W2E (on block 2 by
the Earth).
Taking into account the directions of these forces, we obtain the following free-body
diagrams. (Recall that a free-body diagram doesn’t show the object, but shows the forces
acting drawn from a single point or dot that represents the object.) Here are the diagrams
we come up with:
N1T
+y
T1R
+x
T2R
1
+y
2


W1E
+x
W2E
Note that I decided to choose a coordinate system for block 1 such that the fewest
number of forces do not point along a chosen x or y axis. This will simplify the problem
greatly, but is not required.
2. Find expressions for the x and y components of any forces that are not already in these directions.
The only force on our two free body diagrams that doesn’t point directly along our chosen
axes is the weight force W2E. Notice that the free body diagram for block
1 above already includes these components in grey. The figure at right is a
larger picture of the triangle formed by this vector decomposition,
including the angle.
W1E, y
From the triangle, using the definitions of sine and cosine, we find:
W1E
and
Now we have all our force vectors broken down into components of our
chosen coordinate axes, and we can continue to application of Newton’s
Second Law.
β
W1E, x
PHYS 110 Spring 2011
HW 6.6 SOLUTIONS
3. Write down NII for each block – for both x and y directions.
Newton’s Second Law (NII):
Component Form of NII:
and
For simplicity, we can use shorthand notation for our forces; we can drop most of the
subscripts that tell us the ON and BY information as long as we are still able to distinguish
between the various forces. Before doing this, we should consider the NIII pairs that may
be present since the subscripts can help us with that. We also notice that since the rope is
massless, the tension force on block 1 is the same magnitude as the tension force on block
2.
Here are our simplifications:
T1R = T2R = T
N1I = N
(note that this is the only normal force in this problem)
W1E = W1
W2E = W2
Block 1
(no acceleration in x-direction)
(no acceleration in y-direction)
(1)
(2)
Block 2
(no acceleration in x-direction)
(no acceleration in y-direction)
(no forces in the x-direction)
(3)
PHYS 110 Spring 2011
HW 6.6 SOLUTIONS
4. Assume that you know the following information: β and m1. Find an expression for m2 in terms of
only these two known values.
For this case, we have found 3 equations that give useful information (0 = 0 isn’t very
useful); these are equations (1), (2), and (3) in the boxes above. The question asks for an
expression for m2 in terms of only β and m1; this is essentially a case where the tension T
is unknown. We can combine our 3 equations (actually only two are required here) to solve
the problem. Since T appears in equations (1) and (3), we can solve this system of equations
for the case of two unknowns: m2 and T. There are several ways to do this algebraically;
here is one way:
Solve for T in equation (3):
Substitute our findings into equation (1):
Now we can solve for m2:
If we know β and m1, we could use this relationship to choose m2 to create a situation
where the blocks do not accelerate!
Case II. – Frictionless surface, non-zero acceleration; block 1 moves up the ramp and speeds up.
1. Draw a FBD for each block. Include a coordinate system for each FBD.
The free-body diagrams are essentially the same as in case I, except now the magnitude
of the tension force on block 1 must be greater than the magnitude of the x-component of
the weight force on block 1. Also the magnitude of the weight force on block 2 must be
greater than the magnitude of the tension force on block 2. Notice the subtle difference
in lengths of the force vectors in the free body diagrams below.
PHYS 110 Spring 2011
HW 6.6 SOLUTIONS
N1T
+y
+x
T1R
T2R
+y
1
2

W1E

+x
W2E
2. Find expressions for the x and y components of any forces that are not already in these directions.
These results are identical to case I.
3. Write down NII for each block – for both x and y directions.
The process is similar to case I, but now we have an acceleration! The acceleration of block
1 points up the ramp, and acceleration of block 2 points down. Since the blocks are
connected to a string that doesn’t change length, the accelerations of the two blocks are
the same; let’s call it a.
Block 1
(acceleration in (+) x-direction)
(1)
(no acceleration in y-direction)
(2)
Block 2
(no acceleration in x-direction)
(acceleration in (-) y-direction)
(no forces in the x-direction)
(3)
PHYS 110 Spring 2011
HW 6.6 SOLUTIONS
4. Assume that you know the following information: β, m1, and acceleration a. Find an expression for m2
in terms of only these three known values.
Again for this case, we have found 3 equations that give useful information these are
equations (1), (2), and (3) in the boxes above. The same procedure as case I yields the
following:
Solve for T in equation (3):
Substitute our findings into equation (1):
Now we can solve for m2:
Note that this reduces to our result for case I if we set a = zero!
Case III. – Frictionless surface, non-zero acceleration; block 1 moves up the ramp and slows down.
1.
2.
3.
4.
Draw a FBD for each block. Include a coordinate system for each FBD.
Find expressions for the x and y components of any forces that are not already in these directions.
Write down NII for each block – for both x and y directions.
Assume that you know the following information: β, m1, and acceleration a. Find an expression for m2
in terms of only these three known values.
Result:
PHYS 110 Spring 2011
HW 6.6 SOLUTIONS
Case IV. – Rough surface with µs, zero acceleration. If m2 is any greater, block 1 will begin to slide.
1. Draw a FBD for each block. Include a coordinate system for each FBD.
Following the discussion for #1 from case I, we find the following FBDs:
N1T
1
+y
+x
T2R
+y
T1R
f1I

W1E
2

+x
W2E
2. Find expressions for the x and y components of any forces that are not already in these directions.
These results are identical to case I.
3. Write down NII for each block – for both x and y directions.
The process is similar to case I, except we include a friction force on block 1 that points in the –x
direction.
Block 1
(no acceleration in x-direction)
(no acceleration in y-direction)
(1)
(2)
Block 2
(no acceleration in x-direction)
(no acceleration in y-direction)
(no forces in the x-direction)
(3)
PHYS 110 Spring 2011
HW 6.6 SOLUTIONS
4. Assume that you know the following information: β, m1, and coefficient of static friction µs. Find an
expression for m2 in terms of only these three known values.
In this static case, we have the relationship between the static friction and the normal
force (both by the incline) as
In our case, slipping is about to occur (this is the limiting case), so we can apply the “equals
to” part of the “less than or equals to.”
Notice that information about N is given in equation (2). This time we will combine
information from all 3 equations.
Substituting this result into equation (1), we find:
Solving for T, we find:
Substituting this for T in equation (3), we find:
Solving for m2, we find:
Note that if the coefficient of static friction is zero, we get the result from case I!
PHYS 110 Spring 2011
HW 6.6 SOLUTIONS
Case V. – Rough surface with µk, non-zero acceleration; block 1 moves up the ramp and speeds up.
1. Draw a FBD for each block. Include a coordinate system for each FBD.
The following FBDs are similar to case IV; notice the lengths of the vectors are a little
different:
N1T
+y
1
+y
+x
T2R
T1R
f1I

W1E
2

+x
W2E
2. Find expressions for the x and y components of any forces that are not already in these directions.
These results are identical to case I.
3. Write down NII for each block – for both x and y directions.
Now we have a combination of cases II and IV. Here are the results of applying NII with
the FBDs directly above:
Block 1
(acceleration in (+) x-direction)
(1)
(no acceleration in y-direction)
(2)
Block 2
(no acceleration in x-direction)
(acceleration in (-) y-direction)
(no forces in the x-direction)
(3)
PHYS 110 Spring 2011
HW 6.6 SOLUTIONS
4. Assume that you know the following information: β, m1, and coefficient of kinetic friction µk. Find an
expression for m2 in terms of only these three known values.
For the case of kinetic friction, we have:
The normal force N is found in equation (2), so we have:
Substituting this result into equation (1), we find:
Solving for T, we find:
Substituting this for T in equation (3), we find:
Solving for m2, we find:
Note that if the coefficient of kinetic friction is zero, we get the result from case II!