Part 1:
Q1:
A car is driving down the highway
at 30 m/s. Out of a single thundercloud, a raindrop falls.
The raindrop strikes the side window of the car moving directly
downward at 40 m/s. It glancingly strikes the glass, skimming from
one edge of the glass to the other.
Ignore the effect of drag, gravity and surface tension. Sketch
this situation.
Q2: What
is the velocity (speed and angle) of the raindrop as it moves across
the glass? Describe this mathematically and in your own words.
We know the vertical and horizontal components
of the triangle that is formed when the raindrop moves across the
window. The vertical component is the velocity of the raindrop,
40 m/s. The horizontal component is the velocity of the car, 30
m/s. Employing the Pythagorean theorem, we can solve for the velocity
of the raindrop's path across the window:
c^{2} = a^{2} + b^{2}
c^{2} = (30 m/s)^{2} +
(40 m/s)^{2}
c^{2} = 900 m^{2}/s^{2}
+ 1600 m^{2}/s^{2} = 2500 m^{2}/s^{2}
c = 50 m/s
To find the angle of the path of the raindrop,
we must employ a trigonometric function. To solve for q1,
we can show:
tan q_{1}
= (30m/s)/(40m/s), therefore,
q_{1}
= tan^{1} (30m/s)/(40m/s) = 36.9^{o}
To solve for q_{2},
we can show:
tan q_{2}
= (40m/s)/(30m/s), therefore,
q^{2}
= tan^{1} (40m/s)/(30m/s) = 53.1^{o}
Using this information, we can solve the
problem by saying that the raindrop moves across the window at a
velocity of 50 m/s, and an angle of 53.1^{o} below the negative
xaxis.
Part 2:
Q1:
Determine the magnitude and direction of F_{c} such that
the ring stays stationary. Ignore the force of gravity.
If the ring is to remain stationary, then
the sum of all the forces acting on the ring must be equal and opposite.
To find a solution to this problem, we must break the forces down
into their cartesian components. First, let's deal with F_{a}.
The force F_{a} has a magnitude of 7N in
a direction 23^{o} above the positive xaxis. To find the
x and y components of this force, we must employ some trigonometric
functions. First, to find the y component of F_{a} we can
do as follows:
sin 23^{o} = y/7N, therefore,
y = 7Nsin23^{o} = 2.7N
To find the x component of F_{a},
cos 23^{o} = x/7N, therefore,
x = 7Ncos23^{o} = 6.4 N
Now, let's find the x and y components of F_{b}:
The force F_{b} has a magnitude of 11N
in a direction 40^{o} below the positive xaxis. To find
the x and y components of this force, we must employ some trigonometric
functions. First, to find the y component of F_{b} we can
do as follows:
sin 40^{o} = y/11N, therefore,
y = 11Nsin40^{o} = 7.1 N
To find the x component of F_{b},
cos 40^{o} = x/11N, therefore,
x = 11Ncos40^{o} = 8.4 N
Once again, if the ring is
to remain stationary, then the sum of all the forces acting on the
ring must be equal and opposite. Now we must total the component
forces of F_{a} and F_{b}. The magnitude of F_{c}
must be equal and opposite to te sum of F_{a} anf F_{b}:
F_{a}: 6.4 N x +
2.7 N y
F_{b}: 8.4 N x  7.1 N y (because the
y component falls below the xaxis, it is negative!)
F_{a} + F_{b}
= 6.4 Nx + 2.7 N x + 8.4 Ny  7.1 N y = 9.1 N x + 1.3 N y
We now know that F_{c}
must have the components:  (9.1 N x + 1.3 N y) = 9.1 N X 
1.3 N y
Using the component information
about F_{c}, we can determine the magnitude of F_{c}
and it's direction:
First, to solve for the magnitude of F_{c},
we can employ the Pythagorean Theorem as follows:
c^{2} = a^{2} + b^{2}
c^{2} = (9.1)^{2} + (1.3)^{2}
c^{2} = 84.5
c = 9.2
So, the magnitude of F_{c} is 9.2N
To find the direction of the force F_{c},
we must employ some more trigonometry:
tan q = 1.3/9.1
= 0.14
q = tan1 (0.14) = 8^{o}
Now we know that the force F_{c} has
a magnitude of 9.2N in a direction 8^{o} below the negative
xaxis.
Q2: Sketch
all three forces and describe them in your own words.
The force F_{a} has a magnitude of 7N in
a direction 23^{o} above the positive xaxis. The force
F_{b} has a magnitude of 11N in a direction 40^{o}
below the positive xaxis. The force F_{c} has a magnitude
of 9.2N in a direction 8^{o} below the negative xaxis.
The ring stays stationary because the sum of all three forces is
zero.
