Transcript Forces in Motion

Newtons
nd
2
Forces and kinematics
Law in 2D
Forces on a stationary object
What forces are acting on this stationary object?
Fnormal
mg
Forces on a non-stationary object
If the same object is positioned on an inclined plane, how does that affect
to the two forces shown in the previous slide?
The force due to gravity does not
change direction, but the normal
force acts perpendicular to the
surface it’s on.
Fnormal
ao
mg
Forces on a non-stationary object
What other forces now apply, and why?
The resultant force cause by
gravity which moves the object
down the inclined plane can be
broken into X and Y components
as shown.
Fnormal
mg sin(a)
ao
mg
mg cos(a)
Forces on a non-stationary object
If the ball has a mass of 5kg, the angle is 10o, and its position is 2 meters
up the ramp, then what will be its velocity when it reaches the bottom?
First find the acceleration:
Fdown = 50sin(10) = 5a
a = 1.74m/s2
Fnormal
Then use the other known values to
solve:
s = 2m
u=0
v = 2.64m/s
mg sin(a)
ao
mg
mg cos(a)
Forces on a non-stationary object
A common follow up question on an exam may be something like, if upon
reaching the ground the object experiences a constant frictional
resistance of 2N, how long will it take to come to rest?
Given the resistance we can
calculate the deceleration:
2 = 5a
a = -0.4m/s2
Then use the knowns to solve:
u = 2.64m/s
v=0
t = 6.6s
Fnormal
mg sin(a)
ao
mg
mg cos(a)
Forces being pulled or pushed
This guy is pulling a sled of salamis or fish or something across the ice
at a constant speed. Which force, or forces, is he working against?
Since he is pulling at an angle,
he is working against the
friction in the X direction and
the weight in the Y direction.
Forces being pulled or pushed
This guy is pulling a sled of salamis or fish or something across the ice
at a constant speed. If the sled is 90kg, the angle 30o, and the friction
provides resistance of 45N, then what would be the tension in the rope
if the mass of the rope itself is ignored?
Since friction only acts in the X
direction we can find tension by
45 = Tcos30
T = 52N
Forces being pulled or pushed
This guy is pulling a sled of salamis or fish or something across the ice
at a constant speed. If the sled is 90kg, the angle 30o, and the friction
provides resistance of 45N, then what would be the normal force?
The normal force would no
longer be equal to the weight
since there is a y-component.
The normal can be found by:
mg – Tsin(a) = N
900 – 52sin(30) = 874N
Forces being pulled or pushed
This guy is pulling a sled of salamis or fish or something across the ice
at a constant speed. If the ice becomes more smooth and the friction
drops to 20N, but the man continues to pull with the same force, what
will happen?
The man will accelerate at an
acceleration given by:
The resultant forward force
would be 45 – 20 = 25N
25N = 90a
a = 0.278 m/s2
Forces with pulleys
In this situation the inclined plane is smooth, and the mass of P is great
enough so that it will move down the ramp. What equation could be set
to describe this system with relation to tension T, mass of p mp, gravity
g and acceleration a?
Given that the x-component moves
parallel to the plane, the force
mpg*sin(a) which moves down the
gradient is moving against the
tension T of the string. We also
know that sin(a) = 0.7/2.5.
Therefore:
(0.7/2.5) mpg – T = mpa
Forces with pulleys
In this situation the inclined plane is smooth, and the mass of P is great
enough so that it will move down the ramp. Given that
(0.7/2.5) mpg – T = mpa, what is the tension and the acceleration down
the ramp if the mass of P and Q are 3kg and 0.6kg respectively?
(0.7/2.5) * 3g – T = 3a
T – 0.6g = 0.6a
Solving for a
2.4 = 3.6a
a = (2/3)m/s2
T = 6.4N
Forces with pulleys
In this situation the inclined plane is smooth, and the mass of P is great
enough so that it will move down the ramp. When Q hits the pulley the
string breaks and P moves down the plane, reaching the bottom at a
speed of 2m/s. How long is the string?
First you should find the initial speed,
which will be the speed of Q when it
reaches the pulley.
s=0.7, a=(2/3), u = 0, v = ?
v2 = 2(2/3)(0.7) gives 0.967m/s
Now, acceleration increases to g*sin(a)
which is (0.7/2.5)g or 2.8m/s2
a = 2.8, u = 0.967, v = 2, s = ?
4 = 0.935 + 2 * (2.8) * s
s = 0.55m which means the string is 1.95m