CTN-14. A mass m is accelerates downward along a frictionless inclined plane. The
magnitudes of the forces on the free-body diagram have not been drawn carefully, but the
directions of the forces are correct.
N
a
mg
Which statement below must be true?
A) N < mg B) N > mg
C) N = mg
Answer: N < mg There is no motion and no acceleration along the direction of the
normal force, which is the direction of the y–axis in the tilted coordinate system shown.
Therefore, there must be no net force along that y–axis. So the y–component of the
weight (Fy = –mg cos) must be equal in magnitude to the normal force N. Since mg is
greater than its y–component mg cos, mg must be greater than the normal force N.
y
a
N
 mg
x

A student chooses a tilted coordinate system as shown, and then proceeds to write down
F  m a
equation for the y-direction  F  m a ?
Newton's 2nd Law in the form
x
y
x
,
F
y
y
a
N
 mg
x

A) N  mg sin   m a
B) N  mgcos   ma
y
 m a y . What is the correct
C) mg sin   m a
D) N  mgcos   0
E) N  mg  m a
Answer: N  mgcos   0 (since ay = 0)
CTN-15
A mass m is pulled along a rough table at constant velocity with an external force Fext at
some angle above the horizontal. The magnitudes of the forces on the free-body diagram
have not been drawn carefully, but the directions of the forces are correct.
N
F
ext
F
fric
mg
Which statement below must be true?
A)
B)
C)
D)
E)
Fext > Ffric , N
Fext < Ffric , N
Fext > Ffric , N
Fext < Ffric , N
None of these.
>
<
<
>
mg.
mg.
mg.
mg.
Answer: Fext > Ffric , N < mg . Because velocity is constant, it must be that the net
force is zero. So x– and y–components of forces must cancel. For x: Ffric = Fext,x and
for y: N + Fext,y = mg
N
Ffric
Fext
Fext,x
mg
Fext,y
CTN-16 (Vector review)
The vector A is shown. Axes and positive direction have been chosen as shown. What is
Ax?
A) Ax = A cos
y
B) Ax = A sin

C) Neither of these.
x
Answer: Neither of these! Ax = –A cos (x–component is negative)
CTN-17 A mass slides down a rough inclined plane with some non-zero acceleration a1.
The same mass is shoved up the same incline with a large, brief initial push. As the mass
moves up the incline, its acceleration is a2. How do a1 and a2 compare?
v
v
a2
a1
A) a1 > a2
B) a1 = a2
C) a1 < a2
Answer: a1 < a2 The direction of the friction force is always opposite the velocity. Two
forces affect the acceleration: the frictional force and the component of the weight along
the incline. When these two forces are in the same direction, the net force is large and so
is the acceleration. When these two forces are in opposite directions, the net force is
smaller and the acceleration is smaller.
Ffric
v
v
mg sin

mg sin
Ffric
CTN-18 A block of mass m on a rough table is pulled by a string as shown. The string
exerts a horizontal force of magnitude FT . The coefficient of static friction between
block and table is S; the kinetic friction coefficient is K. As with most surfaces, S >
K .
FT
True (A) or False (B))
If the block does not move, it must be true that the force of friction Ffric = smg .
Answer: False! It MUST be true that Ffric smg. The force of static friction is LESS
THAN OR EQUAL to sN. You only know it is "equal to" if you know the block is just
about to slip.
True (A) or False (B))
If the block does not move, it must be true that the string tension FT  smg .
Answer: True. If the block is not moving, it must be true that Fnet,x = 0, which implies FT
= Ffric . Since it is always true that Ffric sN, it must be true that FT  sNsmg
True (A) or False (B))
If FT > Kmg , the block might be accelerating.
Answer: True
True (A) or False (B))
If FT > Kmg , the block must be accelerating.
Answer: FALSE! Since S > K , it might be that FT > Kmg AND FT < Smg and the
block is stationary because the force of static friction has not yet been overcome.
True (A) or False (B))
If FT > smg , the block must be accelerating.
Answer: True