Problem #1
An ice skater moving at 12 m/s coasts to a halt in 95m on an ice surface. What is the coefficient of (kinetic) friction between ice and skates?Answer:
The information which we are given about the skater’s (one-dimensional) motion is shown
in Fig. 1(a). We know that the skater’s notion is one of constant acceleration so we can use
the results. In particular we know the initial and final velocities of the skater:
v0 = 12 m/s, v = 0
and we know the displacement for this interval:
x − x0 = 95 m
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| Fig.1 |
vx2 = v0x2 + 2ax(x − x0)
ax = (vx2 − v0x2)/[2(x − x0)]
Substituting, we get:
ax = (02 – 122)/[2 x 95] = –0.76 m/s2
Next, think about the forces acting on the skater; these are shown in Fig. 1(b). If the
mass of the skater is m then gravity has magnitude mg and points downward; the ice exerts
a normal force N upward. It also exerts a frictional force fk in a direction opposing the
motion. Since the skater has no motion in the vertical direction, the vertical forces must
sum to zero so that N = mg. Also, since the magnitude of the force of kinetic friction is
given by fk = μkN we have:
fk = μkN = μkmg
So the net force in the x direction is Fx = −μkmg.
Newton’s law tells us: Fx, net = max. Using the results we have found, this gives us:
−μkmg = m(−0.76 m/s2 )
From which the m cancels to give:
μk = [0.76 m/s2]/[9,8 m/s2] = 0.077
The coefficient of kinetic friction between ice and skates is 0.077. (Note, the coefficient
of friction is dimensionless.)
Recall that we were not given the mass of the skater. That didn’t matter, because it
cancelled out of our equations. But we did have to consider it in writing down our equations.
Problem #2
Block B in Fig. 2 weighs 711N. The coefficient of static friction between
block and table is 0.25; assume that the cord between B and the knot is horizontal.
Find the maximum weight of block A for which the system will be stationary.
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| Fig.2 |
Answer:
We need to look at the forces acting at the knot (the junction of the three cables). These
are shown in Fig. 3(a). The vertical cord must have a tension equal to the weight of block
A (which we’ll call WA) because at its other end this cord is pulling up on A so as to support it.
Let the tensions in the other cords be T1 for the horizontal one and T2 for the one that
pulls at 300 above the horizontal. The knot is in equilibrium so the forces acting on it add
to zero. In particular, the vertical components of the forces add to zero, giving:
T2 sin θ − WA = 0 or T2 sin θ = WA
(where θ = 300) and the horizontal forces add to zero, giving:
−T1 + T2 cos θ = 0 or T1 = T2 cos θ
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| Fig.3 |
Now look at the forces acting on the block which rests on the table; these are shown in
Fig. 3(b). There is the force of gravity pointing down, with magnitude WB (that is, the
weight of B, equal to mBg). There is a normal force from the table pointing upward; there
is the force from the cable pointing to the right with magnitude T1, and there is the force of
static friction pointing to the left with magnitude fs. Since the vertical forces add to zero,
we have
N − WB = 0 or N = Wb
The horizontal forces on the block also sum to zero giving
T1 − fs = 0 or T1 = fs
Now, the problem states that the value of WA we’re finding is the maximum value such
that the system is stationary. This means that at the value of WA we’re finding, block B is
just about to slip, so that the friction force fs takes on its maximum value, fs = μsN. Since
we also know that N = WB from the previous equation, we get:
T1 = fs = μsN = μsWB
From before, we had T1 = T2 cos θ, so making this substitution we get
T2 cos θ = μsWB
Almost done! Our very first equation gave T2 = WA/sin θ, so substituting for T2 gives:
(WA/sin θ)cos θ = μsWB or WA cot θ = μsWB
Finally, we get:
WA = μsWB tan θ
Now just plug in the numbers:
WA = (0.25)(711 N) tan 300 = 103 N
Since we solved for WA under the condition that block B was about to slip, this is the
maximum possible value for WA so that the system is stationary.
Problem #3
The two blocks (with m = 16 kg and M = 88 kg) shown in Fig. 4 are not
attached. The coefficient of static friction between the blocks is μs = 0.38, but
the surface beneath M is frictionless. What is the minimum magnitude of the
horizontal force F required to hold m against M?
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| Fig.4 |
Answer:
Having understood the basic set-up of the problem, we immediately begin thinking about
the the forces acting on each mass so that we can draw free–body diagrams. The forces on
mass m are: (1) The force of gravity mg which points downward. (2) The applied force F
which points to the right. (3) The normal force with which block M pushes on m. This
force necessarily points to the left. (4) The frictional force which block M exerts on m. This
is to be a static friction force, so we have to think about its direction... in this case, it must clearly oppose the force of gravity to keep the block m from falling. So we include a force fs
pointing up. These forces are shown in Fig. 5.
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| Fig.5 |
Mg, pointing down; the surface beneath M exerts a normal force N pointing upward. Since
this surface is frictionless, it does not exert a horizontal force on M. The mass m will exert
forces on M and these will be equal and opposite to the forces which M exerts on m. So there
is a force N on mass M pointing to the right and a frictional force fs pointing downward.
Now that we have shown all the forces acting on all the masses we can start to discuss
the accelerations of the masses and apply Newton’s Second Law.
The problem says that mass m is not slipping downward during its motion. This must
mean that the forces of friction and gravity balance:
fs = mg .
But this does us little good until we have an expression for fs. Now, in this problem we are
being asked about a critical condition for the slippage of m. We can reasonably guess that
here the force of static friction takes on its maximum value, namely
fs = μsN ,
N being the normal force between the two surfaces. This is an important bit of information,
because combining that last two equations we get:
mg = μsN .
Let’s consider the horizontal motion of both of the masses. Now, since the masses are
always touching, their displacements, velocities and accelerations are always the same. Let
the x acceleration of both masses be a. Then for mass m, Newton’s Second Law gives us:
∑Fx = F − N = ma
while for mass M, we get
N = Ma
Combining these last two equations gives
F − Ma = ma
F = (M + m)a
a = F/(M + m)
which tells us the force N:
N = Ma = MF/(M + m)
Putting this result for N into our result involving the friction force gives
mg = μsN = μsMF/(M + m)
which lets us solve for F:
F = [(M + m)mg]/(Mμs)
And now we can substitute the given values:
F = [(M + m)mg]/(Mμs)
= [(16 kg + 88 kg)(16 kg)](9.80 m/s2)/[(0.38 (88 kg)]
F = 488N
Problem #4
In Fig. 6 a box of mass m1 = 1.65 kg and a box of mass m2 = 3.30 kg slide
down an inclined plane while attached by a massless rod parallel to the plane.
The angle of incline is θ = 300. The coefficient of kinetic friction between m1 and
the incline is μ1 = 0.226; that between m2 and the incline is μ2 = 0.113. Compute
(a) the tension in the rod and (b) the common acceleration of the two boxes. c)
How would the answers to (a) and (b) change if m2 trailed m1?
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| Fig.6 |
(a) We will shortly be drawing force diagrams for the two masses, but we should first pause
and consider the force which comes from the rod joining the two masses. A “rod” differs from a “cord” in our problems in that it can pull inward on either end or else push outward.
(Strings can only pull inward.) For the purpose of writing down our equations we need to
make some assumption about what is happening and so here I will assume that the rod is
pushing outward with a force of magnitude T, i.e. the rod is compressed. Should it arise in
our solution that we get a negative number for T, all is not lost; we will then know that the
rod is really pulling inward with a force of magnitude |T| and so the rod is being stretched.
With that in mind, we draw a diagram for the forces acting on block 1 and there are a
lot of them, as shown in Fig. 7(a). Rod tension T and the force of kinetic friction on block
1 (to oppose the motion) point up the slope. The “slope” component of gravity m1g sin _
points down the slope. The normal component of gravity m1g cos θ points into the surface
and the normal force N1 from the slope points out of the surface.
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| Fig.7 |
zero, giving:
N1 − m1g cos θ = 0 or N1 = m1g cos θ
The sum of force components in the down–the–slope direction gives m1a, where a is the
down–the–slope acceleration common to both masses. So then:
m1g sin θ − T − fk,1 = m1a
We can substitute for fk,1, since fk,1 = μ1N1 = μ1m1g cos θ. That gives:
m1g sin θ − T − μ1m1g cos θ = m1a
We have a fine equation here, but T and a are both unknown; we need another equation!
The forces acting on block 2 are shown in Fig. 7(b). The force of kinetic friction fk,2
points up the slope. The rod tension T and the “slope” component of gravity m2g sin θ point
down the slope. The normal component of gravity m2g cos θ points into the surface and the
normal force of the surface on 2, N2, points out of the slope.
Again there is no net force perpendicular to the slope, so
N2 − m2g cos θ = 0 or N2 = m2g cos θ .
The sum of the down–the–slope forces on m2 gives m2a, so:
m2g sin θ + T − fk,2 = m2a
We can substitute for the force of kinetic friction here, with fk,2 = μ2N2 = μ2m2g cos θ.
Then:
m2g sin θ + T − μ2m2g cos θ = m2a
Two equations (1 and 2) and tow unknowns (T and a). The physics is done, the rest
is math!
In solving the equations I will go for an analytic (algebraic) solution, then plug in the
numbers at the end. Aside from giving us some good practice with algebra, it will be useful
in answering part (c).
We note that if we add Eqs. 1 and 2, T will be eliminated and we can then find a.
When we do this, we get:
m1g sin θ + m2g sin θ − μ1g cos θ − μ2g cos θ = ma + m2a
Lots of factoring to do here! Pulling out some common factors, this is:
g [(m1 + m2) sin θ − cos θ (μ1m1 + μ2m2)] = (m1 + m2)a
and then we get a:
a = g [(m1 + m2) sin θ − cos θ (μ1m1 + μ2m2)]/(m1 + m2) (3)
But it’s really T we want in part (a). We can eliminate a by multiplying Eq. 1 by m2:
m1m2g sin θ − m2T − μ1m1m2g co
and Eq. 2 by m2:
m1m2g sin θ + m1T − μ2m1m2g co
and then subtracting the second from the first. Some terms cancel, and this gives:
−m2T − m1T − μ1m1m2g cos θ + μ
Factor things:
−T(m1 + m2) = m1m2g cos θ (μ1 − μ2)
and finally get an expression for T:
T = [m1m2g cos θ (μ2 − μ1)]/(m1 + m2) (4)
Hey, that algebra wasn’t so bad, was it? Now we have general expressions for T and a.
Plugging numbers into Eq. 4, we get:
T = [(1.65 kg)(3.30 kg)(9.80 m/s2) cos 300(0.113 − 0.226)]/(1.65 kg + 3.30 kg)
= −1.05 N
Oops! T came out negative! What we find from this is that the assumption was wrong and
the rod is really being stretched as the blocks slide down the slope, and the magnitude of
the rod’s tension is 1.05 N.
(b) To find the acceleration of the blocks, plug numbers into Eq. 3:
a = (9.80 m/s2 )[(1.65 kg + 3.30 kg) sin 300 − cos 300((0.226)(1.65 kg) + (0.113)(3.30 kg)]/(1.65 kg + 3.30 kg)
= 3.62 m/s2
The (common) acceleration of the blocks is 3.62 m/s2 .
(c) Now we ask: What would the answers to (a) and (b) be if the blocks had slid down the
slope with m1 in the lead? Would it make any difference at all? It might, since the friction
forces on the masses come from two different μ’s. In any case, with our analytic results
Eqs. 5.7 and 5.8 we can find the results of switching the labels “1” and “2”, since that is all
we would get from having the blocks switch positions.
If we switch “1” and “2” in Eq. 3, we can see that the result for a will not change at all
because the sums within that expression are not affected by the switch. So the connected
blocks will slide down the slope with the same acceleration, namely 3.62 m/s2 for the given
values.
What about T? From Eq. 5.8 we see that switching “1” and “2” gives an overall change in
sign because of the factor (μ2 −μ1). (The other factors don’t change for this switch.) So we
know that plugging in the numbers for the case where blocks 1 leads would give T = +1.05 N,
and since this is a positive number, the assumption about the rod being compressed (and as
a result pushing outward) would be correct. So for the case where m1 leads, the magnitude
of the rod’s tension is the same (1.05 N) , but now it pushing outward.
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