Problem #5
A 3.0 kg block starts from rest at the top of a 300 incline and slides 2.0 mdown the incline in 1.5 s. Find (a) the magnitude of the acceleration of the block,
(b) the coefficient of kinetic friction between the block and the plane, (c) the
frictional force acting on the block and (d) the speed of the block after it has
slid 2.0m.
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| Fig.8 |
(a) The basic information about the motion of the block is summarized in Fig. 8(a). We
use a coordinate system where x points down the slope and y is perpendicular to the slope.
We’ll put the origin of the coordinate system at the place where the block begins its motion.
The block’s motion down the slope is one of constant acceleration.
(This must be so,
since all of the forces acting on the block are constant.) Of course, this is an acceleration in the x direction, as there is no y motion. It begins its slide starting from rest (v0x = 0) and
so the block’s motion is given by:
x = x0 + v0xt + ½ axt2 = ½ axt2 .
We are told that at t = 1.5 s, x = 2.0 m. Substitute and solve for ax:
2.0 m = ½ ax(1.5 s)2
ax = 2(2.0m)/(1.5 s)2 = 1.78 m/s2
The magnitude of the block’s acceleration is 1.78 m/s2 .
(b) We must now think about the forces which act on the block. They are shown in
Fig. 8(b). Gravity pulls downward with a force mg, which we decompose into its components
along the slope and perpendicular to it. The surface exerts a normal force N. There
is also a force of kinetic friction from the slope. Since the block is moving down the slope,
the friction force must point up the slope.
The block moves only along the x axis; the forces in the y direction must sum to zero.
Referring to Fig. 8(b), we get:
∑Fy = N − mg cos θ = 0
N = mg cos θ.
This gives us the normal force of the surface on the block; here, θ = 300.
The block does have an acceleration in the x direction, which we’ve already found in part
(a). The sum of the forces in the +x direction gives max:
∑Fx = mg sin θ − fk = max
Now we use the formula for the force of kinetic friction: fk = μkN. Using our expression for
the normal force gives us:
fk = μkN = μkmg cos θ
and using this result in the last equation gives
mg sin θ − μkmg cos θ = max .
Here, the only unknown is μk, so we find it with a little algebra: First off, we can cancel the
common factor of m that appears in all terms:
g sin θ − μkg cos θ = ax
and then solve for μk:
μkg cos θ = g sin θ − ax
= (9.80 m/s2) sin 300 − (1.78 m/s2)
= 3.12 m/s2
So we get:
μk = (3.12 m/s2)(9.80 m/s2)(cos 300) = 0.368
(c) As we have seen in part (b), the magnitude of the (kinetic) friction force on the mass is
fk = μkmg cos θ
= (0.368)(3.0 kg)(9.80 m/s2) cos 300
= 9.4N
The force of friction is 9.4N.
(d) We know the acceleration of the block, its initial velocity (v0x = 0) and the time of travel
to slide 2.0 m; its final velocity is
v = v0x + axt = 0 + (1.78 m/s2)(1.50 s) = 2.67 m/s
Problem #6
Three masses are connected on a table as shown in Fig. 9. The table has a
coefficient of sliding friction of 0.35. The three masses are 4.0kg, 1.0kg, and 2.0kg,
respectively and the pulleys are frictionless. (a) Determine the acceleration of
each block and their directions. (b) Determine the tensions in the two cords.
Answer:
(a) First, a little thinking about what we expect to happen. Surely, since the larger mass is
hanging on the left side we expect the 4.0 kg mass to accelerate downward, the 1.0 [kg block
to accelerate to the right and the 2.0 kg block to accelerate upward. Also, since the masses
are connected by strings as shown in the figure, the magnitudes of all three accelerations
must be the same, because in any time interval the magnitudes of their displacements will
always be the same. So each mass will have an acceleration of magnitude a with the direction
appropriate for each mass.
Now we consider the forces acting on each mass. We draw free–body diagrams! If the
tension in the left string is T1 then the forces on the 4.0 kg mass are as shown in Fig. 10(a).
The string tension T1 pulls upward; gravity pulls downward with a force m1g.
The forces acting on m2 are shown in Fig. 10(b). We have more of them to think about;
gravity pulls with a force m2g downward. The table pushes upward with a normal force N.
It also exerts a frictional force on m2 which opposes its motion. Since we think we know
which way m2 is going to go (left!), the friction force fk must point to the right. There are
also forces from the strings. There is a force T1 to the left from the tension in the first string
and a force T2 pointing to the right from the tension in the other string. (Note, since these
are two different pieces of string, they can have different tensions.)
The forces on m3 are shown in Fig. 10(c). There is a string tension T2 pulling up and
gravity m3g pulling down.
All right, lets write down some equations! By Newton’s Second Law, the sum of the downward
forces on m1 should give m1a. (We agreed that its acceleration would be downward.)
This gives:
m1g − T1 = m1a (5)
Moving on to mass m2, the vertical forces must cancel, giving
N = m2g .
Newton tells us that the sum of the left–pointing forces must give m2a (we decided that its
acceleration would be of magnitude a, toward the left) and this gives:
(we decided that its acceleration would be of magnitude a, toward the left) and this gives:
T1 − fk − T2 = m2a
But since
fk = μkN = μkm2g ,
this becomes
T1 − μkm2g − T2 = m2a (6)
Finally, the sum of the upward forces on m3 must give m3a. So:
T2 − m3g = m3a (7)
Having done this work in writing down these wonderful equations we stand back, admire
our work and ask if we can go on to solve them. We note that there are three unknowns (a,
T1 and T2) and we have three equations. We can find a solution. The physics is done. . . only
the algebra remains.
We can do the algebra in the following way: If we just add Eqs. 5, 6 and 7 together
(that is, add all the left–hand–sides together and the right–hand–sides together) we find that
both T’s cancel out. We get:
m1g − T1 + T1 − μkm2g − T2 + T
which simplifies to:
m1g − μkm2g − m3g = (m1 + m2 + m3)a
Now we can easily find a:
a = (m1 − μkm2 − m3)g/(m1 + m2 + m
= [(4.0kg) − (0.35)(1.0kg) − (2.0 kg)](9.80 m/s2)/(4.0kg +1.0kg +2.0kg)
= (1.65 kg)(9.80 m/s2)/(7.0kg)
= 2.31 m/s2
So our complete answer to part (a) is: m1 accelerates at 2.31 m/s2 downward; m2 accelerates
at 2.31 m/s2 to the left; m3 accelerates at 2.31 m/s2 upward.
(b) Finding the tensions in the strings is now easy; just use the equations we found in part
(a).
To get T1, we can use Eq. 5, which gives us:
T1 = m1g − m1a = m1(g − a)
= (4.0 kg)(9.80 m/s2 − 2.31 m/s2)
= 30.0N .
To get T2 we can use Eq. 7 which gives us:
T2 = m3g + m3a = m3(g + a)
= (2.0 kg)(9.80 m/s2 + 2.31 m/s2)
= 24.2N .
The tension in the string on the left is 30.0 N. The tension in the string on the right is
24.2N.
Problem #7
A block is placed on a plane inclined at 350 relative to the horizontal. If the
block slides down the plane with an acceleration of magnitude g/3, determine
the coefficient of kinetic friction between block and plane.
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| Fig.11 |
The forces acting on the block (which has mass m) as it slides down the inclined plane are
shown in Fig.11. The force of gravity has magnitude mg and points straight down; here
it is split into components normal to the slope and down the slope, which have magnitudes
mg cos θ and mg sin θ, respectively, with θ = 350. The surface exerts a normal force N and a force of kinetic friction, fk, which, since the block is moving down the slope, points up the
slope.
The block can only accelerate along the direction of the slope, so the forces perpendicular
to the slope must add to zero. This gives us:
N − mg cos θ = 0
N = mg cos θ
The acceleration of the block down the slope was given to us as a = g/3. Then summing
the forces which point along the slope, we have
mg sin θ − fk = ma = mg/3
The force of kinetic friction is equal to μkN, and using our expression for N we have
fk = μkN = μkmg cos θ
and putting this into the previous equation gives:
mg sin θ − μkmg cos θ = mg/3 .
Fortunately, the mass m cancels from this equation; we get:
g sin θ − μkg cos θ = g/3
And now the only unknown is μk which we solve for:
μkg cos θ = g sin θ – g/3 = g(sin θ – 1/3)
Here we see that g also cancels, although we always knew the value of g! We then get:
μk = (sin θ – 1/3)/cos θ
= (sin 350 – 1/3)/cos 350
μk = 0.293
So the coefficient of kinetic friction between block and slope is 0.293.
Problem #8
A 2.0 kg block is placed on top of a 5.0 kg as shown in Fig.12. The coefficient
of kinetic friction between the 5.0kg block and the surface is 0.20. A horizontal
force F is applied to the 5.0kg block. (a) Draw a free–body diagram for each
block. What force accelerates the 2.0kg block? (b) Calculate the magnitude of
the force necessary to pull both blocks to the right with an acceleration of 3.0 m/s2.
(c) find the minimum coefficient of static friction between the blocks such that
the 2.0kg block does not slip under an acceleration of 3.0 m/s2 .
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| Fig.12 |
(a) What forces act on each block?
On the big block (with mass M = 5.0 kg, let’s say) we have the applied force F which
pulls to the right. There is the force of gravity, Mg downward. The surface exerts a normal
force N1 upward. There is a friction force from the surface, which is directed to the left. The small mass will also exert forces on mass M; it exerts a normal force N2 which is directed
downward; we know this because M is pushing upward on m. Now, M is exerting a force
of static friction fs on m which goes to the right; so m must exert a friction force fs on M
which points to the left.
These forces are diagrammed in Fig. 13(a).
On the small block we have the force of gravity, mg downward. Mass M exerts an upward
normal force N2 on it, and also a force of static friction fs on it, pointing to the right. It
is this force which accelerates m as it moves along with M (without slipping). These forces
are diagrammed in Fig. 13(b).
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| Fig.13 |
Notice how the forces between M and m, namely N2 (normal) and fs, have the same
magnitude but opposite directions, in accordance with Newton’s Third Law. They are so–
called “action–reaction pairs”.
(b) The blocks will have a horizontal acceleration but no vertical motion, so that allows us
to solve for some of the forces explained in part (a). The vertical forces on m must sum to
zero, giving us:
N2 − mg = 0
N2 = mg = (2.0 kg)(9.80 m/s2) = 19.6 N
and the vertical forces on M must sum to zero, giving us:
N1 − N2 −Mg = 0
N1 = N2 + Mg = 19.6 N + (5.0 kg)(9.80 m/s2)
= 68.6N
We are given that the acceleration of both blocks is 3.0 m/s2 . Applying Newton’s Second
Law to mass m we find:
∑Fx = fs = max = (2.0 kg)(3.0 m/s2) = 6.0 N
While applying it to M gives
∑Fx = F − fk − fs = Max = (5.0 kg)(3.0 m/s2)
= 15.0N
We found fs above; we do know the force of kinetic friction (from M’s sliding on the surface)
because we know the coefficient of kinetic friction and the normal force N1:
fk = μkN1 = (0.20)(68.6 N) = 13.7 N
Now we can solve for F:
F = 15.0 N + fk + fs
= 15.0 N +13.7 N +6.0 N
= 34.7 N
To pull the blocks together to the right with an acceleration 3.0 m/s2 we need an applied force
of 34.7 N.
(c) As we’ve seen, mass m accelerates because of the friction force fs (from M’s surface)
which acts on it. Forces of static friction have a maximum value; here we know that we must
have
fs ≤ μsN2
in order for m not to slip on M. Here, we have fs = 6.0N and N2 = 19.6 N. So the critical
value of μs for our example is
μs = f2/N2 = 0.306
If μs is less than this value, static friction cannot supply the force needed to accelerate m
at 3.0 m/s2. So μs = 0.306 is the minimum value of the coefficient of static friction so that the
upper block does not slip.
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