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Newton’s Second Law
F = ma
Solving problems when  F = ma
 F  ma
Newton’s Second Law
F = ma
The cyclist has a mass of 50 kg and is accelerating at 0.9 m/s2.
What is the size of the unbalanced force acting on the cyclist?
Newton’s Second Law
F = ma
How to approach a dynamic problem:
• Draw a free-body diagram
• Choose a coordinate axis and resolve all forces into
components.
• Set the sum of the force components each equal to ma.
• Solve the resulting equations for the unknowns.
Newton’s Second Law
F = ma
1) Read the problem.
(identify givens, look for “hidden” knowledge)
2) Draw a free-body diagram
(identify all forces acting upon object)
3) Add all forces in one direction together (x?)
 F = F1 + F2 + F3 + …
(determine sum of forces, maybe Fnet = 0 or Fnet = ma)
4) Add all forces in other direction together (y?)
(determine sum of forces, maybe Fnet = 0 or Fnet = ma)
5) Solve for what you don’t know
Newton’s Second Law
F = ma
If each person is pushing forward against the 1,500 N car, find:
a. The Normal force
b. The acceleration of the car
Newton’s Second Law
F = ma
If the 10 Newton crate is being pushed forward by a
force P of magnitude 80 N at an angle of 300 as shown
find:
1. The mass of the crate
2. The Normal force
3. The acceleration of the crate
Newton’s Second Law
F = ma
If m1=20 kg and m2=70 kg find:
1. The tension in the cable
2. The acceleration of the masses
Newton’s Second Law
F = ma
600N
Given: The car is accelerating
forward at 2m/s2 and  = 25°
1m
Find:
The forces in the ropes AB
and AC.
Newton’s Second Law
F = ma
A stream of water strikes a stationary turbine blade, as the drawing
illustrates. The incident water stream has a velocity of +18.0 m/s,
while the exiting water stream has a velocity of -18.0 m/s. The mass
of water per second that strikes the blade is 25.0 kg/s. Find the
magnitude of the average force exerted on the water by the blade.
Newton’s Second Law
F = ma
Robin Hood (m =82 kg) is escaping from a dangerous situation.
With one hand he is gripping the rope that holds up a chandelier
(m =220 kg). When he cuts the rope where it is tied to the floor, the
chandelier will fall, and he will be pulled up to the balcony. Find:
1. the acceleration with which Robin is pulled upward
2. the tension in the rope while Robin escapes.
Newton’s Second Law
F = ma
 What is the net force
acting on the mule?
 What is the approximate
answer you expect to get?
 Begin by calculating the
components of each force.
Newton’s Second Law
F = ma
F1x  120 N  ? 60
F1x  120 N  cos 60
 60 N
F2 x  20.7 N
R  185 N
Rx  39.3 N
@12.3 right
F1 y  120 N  sin 60
 104 N
F2 y  77.3 N
Ry  181 N
Newton’s Second Law
F = ma
A 3.0 kg mass hangs at one end of a rope that is attached
to a support on a railroad car. When the car accelerates to
the right, the rope makes an angle of 4.0° with the
vertical. Find the acceleration of the car.
Newton’s Second Law
F = ma
In the vertical direction
In the horizontal direction
Newton’s Second Law
F = ma
A block S (the sliding block) has a mass M = 3.3 kg. The block is
free to move along a horizontal frictionless surface and is
connected by a cord that wraps over a frictionless pulley to a
second block H (the hanging block), with mass m = 2.1 kg. The
cord and pulley are “massless”. The hanging block H falls as the
sliding block S accelerates to the right.
Newton’s Second Law
F = ma
An inclined plane making an angle of 25o with the horizontal has a
pulley at its top. A 30 kg block on the plane is connected to a freely
hanging 20 kg block by means of a cord passing over the pulley.
a. Compute the distance that the 20 kg block will fall in 2.0 seconds
starting from rest. Neglect friction.
b. We now cut the cord. As the block then slides down the inclined
plane, does it accelerate? If so, what is its acceleration?
Newton’s Second Law
F = ma
 F  T  mg sin   ma
T  mg sin 
 (15 kg ) (9.8 m / s 2 )(sin 27 )
 67 N
 F  N  mg cos   0
N  mg cos 
 (15 kg ) (9.8 m / s 2 )(cos 27 )
 131 N
T N w0
mg  sin   ma
a  g  sin 
Positive direction is
down the inclined plane
a  (9.8m / s 2 ) (sin 27 )
 4.4 m / s 2
Given:
M1 = 12.0 kg
Newton’s Second Law
F = ma
M2 = 24.0 kg
M3 = 31.0 kg
T3 = 65.0 N
What is the tension T2
T2  T1  m2 a
T2  m2 a  T1
T2  (24.0)(0.97)  (11.64) N
T2  34.92 N
T2  (m1  m2 )a
T2  (12.0  24.0)(0.97) N
T2  34.92 N