PY1001 Problem Set 3 – Solutions
N
θ
Fw
θ
θ
mg
(1) A 45-kg skier skies down a frictionless ski slope inclined at an angle
of 15◦ to the horizontal. (a) What is the skier’s acceleration down the
slope? (b) What is the normal force on the skier due to the slope? Suppose
that when a strong wind blows horizontally against her as she skies down
her acceleration down the slope is 1.2 m/s2 . (c) What is magnitude of the
force of the wind? (d) What is the normal force due to the slope in this
case? Make sure to draw diagrams showing the forces acting and how they
break up into components in both cases.
(a) The only force with a component down the slope is the skier’s weight. The
component of this force down the slope is
Fx = mg sin θ −→ ax = g sin θ
ax = (9.8 m/s2 ) sin(15.0◦ ) = 2.54 m/s2
(b) The sum of the forces perpendicular to the slope must be zero:
N − mg cos θ = 0 −→ N = mg cos θ
N = (45.0 kg)(9.8 ms/2 ) cos(15.0◦ ) = 426 N
(c) After breaking up the force of the wind Fw into components up the slope and
perpendicular to the slope, we now have along the slope
sin θ−a)
mg sin θ − Fw cos θ = ma −→ Fw = m(gcos
θ
Fw =
(45.0 kg)(9.8 sin(15◦ −1.2) m/s2
cos(15◦ )
= 62 N
(d) We have in the direction perpendicular to the slope, with the sum of the forces
still being zero (no acceleration in this direction):
N − mg cos θ − Fw sin θ = 0 −→ N = mg cos θ + Fw sin θ
N = (45.0 kg)(9.8 ms/2 ) cos(15.0◦ ) + (62 N) sin(15◦ ) = 442 N
(2) A horse is used to pull a barge along a canal. The horse pulls on the
rope with a force of 7500 N at an angle of 20◦ to the direction of the canal.
The mass of the barge is 9200 kg, and it moves straight along the canal
with an acceleration of 0.10 m/s2 . What are the magnitude and direction
of the force on the barge by the water?
Here, we have two forces acting on the barge: the force due to the horse F H and the
force due to the water FW . The sum of these forces must point in the direction of
motion of the barge, straight along the canal, and the magnitude of the sum of these
forces must be the barge’s mass times its acceleration. Look at the forces in the x
and y directions, where the +x direction is in the direction of the motion of the barge
and the +y direction is toward the bank where the horse is walking:
ΣFx = FHx + FW x = FH cos(20◦ ) + FW x = mB aB
ΣFy = FHy + FW y = FH sin(20◦ ) + FW y = 0
Now we can find the x and y components of the force on the barge due to the
water:
FW x = mB aB − FH cos(20◦ )
= (9200 kg)(0.10 m/s2 ) − (7500 N)(0.940)
FW x = −6130 N
FW y = −FH sin(20◦ )
= −(7500 N)(0.342)
FW y = −2565 N
The magnitude and direction of FW are given by
q
2
2
FW =
FW
x + FW y =
FW = 6645 N
q
(−6130 N)2 + (−2565 N)2
θW = arctan(−2565 N)(−6130 N) = 203◦
The direction of F~W is 203◦ = −157◦ from the +x direction.
(3) A 110-gram hockey puck sent sliding over ice with an initial speed
of 6.0 m/s is stopped in 15 m by the frictional force on it from the ice.
(a) What is the acceleration experienced by the puck? (b) What is the
magnitude of the frictional force on the ice? (c) What is the coefficient of
kinetic friction between the puck and the ice?
(a) Since the frictional force is constant, the acceleration will be constant, and we can
use our equations for constant acceleration. We know the initial velocity (6.0 m/s),
the final velocity (0 m/s) and the distance travelled (15 m). Therefore,
v 2 = vo2 + 2a(x − xo )
a=
v 2 −vo2
2(x−xo )
a=
02 −6.0 m/s2
2(15 m)
= −1.2 m/s2
(b) The magnitude of the frictional force will be
Fk = m|a| = (0.110 kg)(1.2 m/s2 ) = 0.132 N
(c) The frictional force is given by
F = µk N
In the vertical direction, there are two forces, the puck’s weight and the normal force,
which must add to zero:
N − mg = 0 −→ N = mg
Therefore,
F = µk N = µk mg
F
µk = mg
= ma
= ag
mg
µk =
1.2 m/s2
9.8 m/s2
= 0.12