The
Newton’s second law : Force (vector ) = mass (scalar) times
the acceleration (vector)
We
now work in 2 D. F = m a becomes 2 equations. 1 for each of the 2
directions (X and Y)
Fx
(sum
of all the x-components of the forces acting on the body
) = max
(acceleration
along the horizontal)
Fy
(sum
of all the x-components of the forces acting on the body
) = may
(acceleration
along the horizontal)
EXERCICES:
0) write the following on index cards:| REMEMBER: to find an adjacent side, use cosine and multiply by the hypothenus To find the opposite side, use sine and multiply by the hypothenus The hypothenus is the magnitude of the vector you are observing. If the vector is the weight , the hypothenus is mg To find the direction of a vector, first trace it (to identify the quadrant), then use tan( θ ) = Fy / Fx down = negative component, left = negative component |
|
| ALONG THE X-AXIS | ALONG THE Y-AXIS |
| first: Write all the X-components Px = ------- (x-component of engine force if any) Nx = 0 (normal force) Wx= -------- (if any x-component of the weight) Tx = ------------ (x-component of the tension if any) fx = ------------- (friction if any) ..... |
first: Write all the Y-components Ny = ---------- (normal) Py = ------------ (y-component of engine) Ty = ------------ (y-component of tension) Wy = --------(y-component of the weight) fy = 0 (usually, no friction along the y-axis) ........ |
| second: Sum all the X-components = Net force acting along the X-axis = Fnet x | Sum all the Y-components = Fnet y |
| third: Fnet x = m ax if the object is accelerating along the X-axis, ax is the acceleration, m the mass or Fnet x = 0 if the object is moving at a constant speed or is not moving at all. ax = 0 |
Usually the object is not accelerating along the y-axis so: Fnet y = 0 because ay = 0 (if this is not the case, Fnet y = m ay ay is the acceleration along the y-axis) |
1)What
thrust (P) is
needed to fire a 350kg rocket straight up with an acceleration of 8.0
m/s2 ?
Hint: ay = 8m/s/s , W = - mg with g = 9.8 m/s/s , and P +
W = m ay along vertical
Hint: P = m ax
solve for ax
4) 2 forces of 50N
(F1) and the other of 30N (F2) , act in opposite direction on a box as
shown in the diagram.
What is the mass of the box if its acceleration is 4.0 m/s2 ? (see diagram right)
Hint: F1x = 50N ,
F2x = - 30N and
F1x+F2x=max
5)
A 6 kg block being pushed across a table by a force P has an acceleration
of 3.0 m/s2
A) what is the net force
acting on the block. (hint: use Fnet = m a)
B) If the magnitude of P is 20N, what is the
magnitude of the frictional force acting on the block ?
(hint: write Px + Fx = max
solve for Fx. Must be negative)
A) What is the net horizontal force acting on the block ?
(hint: Find the
x-components of each force and write Fnetx = F1x + F2x +F3x
One
component is negative: F1x = -10 N, F2x = 5N and F3x = 25N)
B) What is the
horizontal acceleration of the block?
Hint: Fnet = m ax
7)
what is the weight of a 40kg mass ? (use acceleration due to gravity.
Weight is the pull of gravity)
8)
At a given instant in time, a 4kg rock that has been dropped from a
high cliff experiences a force of air resistance of 15N. What are the
magnitude and direction of the acceleration of the rock. (do not forget
the force of gravity)
Hint: make a
diagram. Along the vertical the air resistance F points up and the weight W points down. Write Fy + W = m ay
9) An upward force of
18N (F) applied via a string to lift a ball with a mass of 1.5 kg.
A) what is the net force
acting upon the ball ? (
Hint: Fnet = W
+ F W<0
A) What is the
acceleration of the ball ?
Hints: write W + F = m ay
10) A child is pulling
a sled (with his friend on it ) and the sled accelerates.
What is the horizontal acceleration ax of the child on the sled, with
combined mass 40kg, if the friction is 60N (F in red and the force is
being pulled with a force of 170N at an angle of 35 degrees with the
ground (P = 170N @ 35 in green)
? See diagram.
Hints: Find the x-component of each force first:
Px = __________ (use cosine)
Fx = ____________ (negative value, friction)
Then write: Px + Fx = m ax and solve for ax.
11)
A 2.0kg weather balloon is released and begins to rise against 6.5 N of
viscous drag (D). If its buoyancy is 32N (B), What is its acceleration ?
Hints : buoyancy
is up B, viscous drag is down D, weight is down W
Write By + Dy + W = m ay
with Vy <0 and W <0
solve for ay
11)
A 2.0kg weather balloon is released and begins to rise against 6.5 N of
viscous drag (D). If its buoyancy is 32N (B), What is its acceleration ?
Hints : buoyancy is up B, viscous drag is down D, weight is down W
Write By + Dy + W = m ay with
Vy <0 and W <0 solve
for ay
12)
A block (5kg) is moving down an inclined plane. The inclined plane
makes a 20 degrees (A) angle with the
horizontal. The forces are in red and the components of forces, if any, are in blue The coordinate
system (x,y) is in orange.
Try without hints. (see diagram)
A) Can you label the
vectors ?
Find: the weight W of the block (always
pointing down)
The x-component of the weight Wx
The y-component Wy
The normal force N (recoil of the plane,
pushing up)
B) write sum of
x-components = m ax or Wx = m ax
and find the acceleration ax along the block.
Hints: Wx = mg sin (A)
and solve for ax
C) Find the
normal force.
Hints: along the
vertical, there is no acceleration so sum of y-components = 0
So Wy + N = 0
with Wy = -mg cos(A) (the component is negative)
Substitute Wy in Wy + N = 0 and solve for N.
D) optional.
(for Physics lovers)
There is now a friction force F = 10N along the plane. Draw the diagram
again and find the new value of the acceleration. Find the coefficient
of friction µk. (F = µk N)
13) A sky diver is falling from a aeroplane.
Hints: what
force pull you toward the earth ? Air above the earth will produce ...
B) State how
each force changes as the sky diver speeds down
Hints: do you
think the weight vary ? What about air resistance?
C) Why does
the sky diver reach a steady speed ? (terminal speed)
D) Describe and explain what happen when the sky diver opens the
parachute.
Hints: suddenly the air drag
will become larger than the weight
14) for physics lovers. Optional.
A 2.5 kg block slides down a 25 degrees inclined place with constant acceleration.
The block starts from rest at the top. At the bottom, its velocity is 0.65 m/s. The incline is 1.6 m long.
hint:
For question 1) you need to use Kinematics to solve for the acceleration a (a vector)
------------------------------------------------------------------------------------------------------------------------------------------------------------
Here are the 3 equations of motion you need to know to survive in Kinematics:
1) d = Vi t + 0.5 a t**2 (d is the displacement, a the acceleration, Vi the initial velocity (0 if it starts at rest) and t the time)
2) a = ( V – Vi ) /t (V is the final velocity (when the time t elapses), Vi is the initial velocity)
3) V**2 = Vi **2 + 2 a d (V**2 meaning V final squared)
V, Vi, a and d are vectors. Can be positive or negative.
a is positive if the object is accelerating when moving toward the + direction
a is negative if the object is decelerating when moving toward the + direction
V is positive if the object is moving toward the + direction
B) What is the coefficient of friction
C) Does the result of either A or B depend on the mass of the block ?
15) A pendulum swings. The diagram (right) shows the pendulum at a given time of its motion.
A) Label on the diagram:
W the weight of the mass
Wx the x-component of W
Wy the yc0mponent of W
T the tension the string pulling back on the mass
B) optional
Find ax using g and θ (use cosine or sine)
Use m ax = Wx
C) If θ
is small,
suppose sin (θ)
= θ
In that case, ax =
___________
D) Find T using m, g, and θ
(use
FIREFOX)
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