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a
Poor Man’s Accelerometer
A driver hangs an air freshener from their rearview mirror with a string. When accelerating onto the highway, the driver notices that the air freshener makes an angle of about 25 degrees with respect to the horizontal. What is the acceleration of the car?
(1) Comprehend the Problem We’ve got a hanging air freshener (of unknown mass) accelerating forward. It’s hanging from a string at 25 degrees from the vertical. From this angle we’re to find the object’s acceleration. Since the air freshener appears at rest to the driver, it must be accelerating at the same rate as the car. Since the car’s acceleration is all horizontal, the air freshener’s acceleration should be entirely horizontal as well.
(2) Express the Problem in Formal Terms (Describe the Physics) We’re looking for the acceleration of the air freshener, so we probably need to find all the forces acting on the air freshener that cause this acceleration. The free body diagram is shown at the right.
There are two forces on the air freshener, a tension up and to the left and a weight straight down. The acceleration is to the left, so the sum of these two forces must point to the left. We’ll choose to the left as the +x direction so the acceleration is purely x (this should make the math simpler later).
The relationship between forces and acceleration is given by Newton’s
Second Law, Fnet =ma
. To find the net force, we’ll need to know the
directions of all the forces in the problem. We know the weight points straight down (all y-direction), and the tension in the string has a positive x-component and a positive y-component (i.e. “up and to the left”).
(3) Plan a Solution We’ll write Newton’s Second Law as two component equations, one in the x-direction and the other in
the y-direction. We’ll have two unknowns, the magnitude of the tension T and the acceleration in the
x-direction ax. We can solve the two equations for the two unknowns T and ax.
T
W
x
y
(4) Execute the Plan
Write Newton’s Second Law as two component equations, one in the x- direction and the other in the y- direction.
net
net, , x
net, ,
ma
ma
i i
x i x i y i y y i
F F ma
F F
F F
=^ =
Express the net force as a sum of all the forces acting on the object.
x-direction:
( 25 ) (^0 )
net,
sin
sin
x x x x x
x
x
F ma T W ma
T ma
T ma
y-direction:
( 25 ) (^ )^ (^0 )
net,
cos
cos
cos
y y
y y y
F ma
T W ma
T mg m
T mg mg T
We can insert the tension value from the y-direction equation into the x- direction equation. (^) ( )( )
2
2
m s m s
sin
sin cos
tan..
.
x
x
x
x
T ma mg ma
a g
a
The air freshener (and therefore the car) accelerates at 4.6 m/s^2.
T
W
x
y
For both questions a) and b), we can identify the forces on the object and are asked about the acceleration. The relationship between forces and acceleration is embodied in Newton’s Second Law:
where the net force is the sum of all the forces acting on the crate. This vector equation stands for two scalar (i.e. “just number” ) equations, one in the x-direction and one in the y-direction:
where each equation deals only with x- or y-components, respectively. We know the following parameters:
(3) Plan a Solution
a) We know the magnitude of the tension in the cable and we can find the magnitude of the crate’s weight using the relationship between weight and the acceleration due to gravity near earth,.
We also know the angle the tension makes with the horizontal, since it points along the cable at 15 degrees. To find the initial acceleration of the crate, we can find the vector sum of all the forces on the crate and divide by the mass. This will give us the initial acceleration. b) Here we are asked to find the first cable angle that lifts the crate off of the ground. This happens just after the normal force goes to zero. How do we know this? The normal force is needed initially because the vertical component of the cable tension is not as big as the vertical component of the crate’s weight downward. (see the figures on the next page) This means that if there were no normal force, there would be a net force downward on the crate. The crate would therefore accelerate downward and penetrate the floor. Since this doesn’t happen, the floor must be pushing upward with a normal force just big enough to make up the difference between the y-component of the tension and the weight. As the crate approaches the tower, the cable points in a more vertical direction. This increases the y-component of the tension. Eventually the tension’s y-component will be large enough to cancel the crate’s weight on its own, and the normal force won’t be needed (i.e. it will be zero). If the y-component of the tension grows past this value, there will be a net force upward on the crate (the normal force can’t pull downward unless the floor is “sticky”). A net upward force means that the crate will accelerate upward.
Initial Situation: A normal force is needed to keep the block from accelerating downward
Later: The tension’s y-component just cancels the crate’s weight
Even Later: The tension’s y-component is larger than the crate’s weight
N
W
T
T y
T x
Free Body Diagram
W
N
T y
y component s
W
T
T y
T x
Free Body Diagram
T y
W
y component s
W
T
T y
T x
Free Body Diagra m
T y
W
y component s
(no y-acceleration) (no y-acceleration) (accelerates upward)
We should therefore write the equation for the net force on the crate in the y-direction. We will then set the normal force magnitude equal to zero and solve for the angle that makes this happen. That will be the last angle at which the crate remains on the floor. Any larger angle will accelerate the crate upward.
(4) Execute the Plan
a) Find the magnitude of the crate’s initial acceleration from Newton’s Second Law. Break the general statement of Newton’s Second Law into x- and y-component equations.
We are told that the crate initially slides along the floor, so its y-velocity remains zero. This implies that its y-acceleration is zero. All that’s left is to find the x-acceleration.
^ ( )
net,^00
net,
net
F ma m
F ma
F ma
y y
x x
Write the net force on the crate as a sum of all the forces acting on the crate (see the free body diagram drawn earlier). We’ve chosen horizontal and vertical as our x- and y-axes. Up and right are the positive directions. We’ll stick with the x- direction since we already know that the y- acceleration is zero.
( ) x
x x x x
x x
T ma
N W T ma
F ma
( 0 ) ( 0 ) cos( 0 )
net,
θ
Solve for the x-component of the acceleration and insert the numbers from the problem.
2
2 s
s m
kgm
0
0
100 kg
3000 cos 15
cos
cos
m
T
a
T ma
x
x θ
θ
Groovin’
A sphere of mass M is resting in the groove created by two inclined planes meeting at their tips. The planes have known different angles of incline with respect to the horizontal. Find the magnitude of the normal force supplied by each plane.
(1) Comprehend the Problem We have a sphere resting on two flat surfaces, each with its own known incline with the horizontal. A sketch of the situation looks like this:
θ 1 θ 2 θ 1 θ 2
Ν 1
Ν 2
W
We know that the sphere is resting on the incline. We know the mass of the ball, as well. We are also assumed to know the two angles of the incline, θ 1 and θ 2. We know that there is a gravitational force (the weight) pulling the sphere straight down. Therefore, there must also be a normal force on the ball from each plane to keep the ball from penetrating the surfaces.
We are to find the magnitude of each of these normal forces. Since we know the ball remains at rest, its acceleration must be zero. We therefore know its acceleration must be zero (velocity never changes from zero). Since we know the acceleration and are asked about forces, we’ll probably use Newton’s Second Law to determine the normal forces.
(2) Represent the Problem in Formal Terms (Describe the Physics) We can organize our qualitative idea about what’s going on with a free body diagram of the ball. This means that we draw the ball by itself (i.e. “free” of the rest of the objects) with only the forces ON the ball.
We know we have three forces to deal with. We’ll probably need to know their directions. We can choose our x- and y-axes any way we like, but since none of these forces are either perpendicular or parallel to each other we might as well just choose up as +y and right as +x. This at least makes the weight force point entirely in the y-direction.
To find the direction of the normal forces, we’ll need to use some geometry. Let’s look at N 1 first:
The normal force is always directed perpendicular (i.e. “normal”) to the surface that provides it. If we extend N1’s direction back to the surface that provides it, we meet that surface at a right angle, as shown.
Extending this line all the way to the x-axis makes a right triangle underneath the left-hand surface. All the angles of a triangle add up to 180 o, so the angle we’re interested in should be: 180 o^ – 90o^ – θ 1 = (90o^ – θ 1 ) Similarly, the angle between N 2 and the horizontal is (90o^ – θ 2 ). Note that N 2 points in the negative x-direction while N 1 points in the positive x-direction.
Ν 1 Ν 2
W
x
y
θ 1 θ 2
90 ο−θ 1
Ν 1
We therefore have the following forces:
1 1 1 2 2 2 90
magnitude normal force from the left surface angle from the horiztonal
magnitude normal force from the right surface angle from the horiztonal
weight of the sph
N
N
N
N
W
θ
θ
magnitude mg ere angle - from the horiztonal
We can relate these forces by using Newton’s Second Law:
net i i
F = ∑F =ma
(3) Plan a Solution We will break Newton’s Second Law into x- and y-component equations. Since we know the acceleration is zero (sphere remains at rest), we can solve these two equations for the two unknowns N 1 and N 2.
(4) Execute the Plan
First, we break Newton’s Second Law down into two component equations.
We know that the sphere remains at rest, so the acceleration along both the x- and y-directions is
zero. (^ )
net
net, ,
net, ,
net, ,
net, ,
ma
ma
m
m
i i
x i x x i y i y y i
x i x i y i y i
F F ma
F F
F F
F F
F F
=^ =
=^ =^ =
Now we write the net force in each direction as the sum of each force’s components. (see the free body diagram in step (2))
We are also using the trigonometric identities
( ) (^ )
( ) (^ )
sin cos
cos sin
θ θ
θ θ
x-direction
( (^ )) ( (^ )) (^ )
1 2
1 1 2
1 1 2 2
net,
cos cos
sin sin
x x x x
F N N W
N N
N N
θ θ
θ θ
y-direction
( ( )) ( ( )) (^ )
1 2
1 1 2
1 1 2 2
net,
sin sin
cos cos
y
y y y
F
N N W
N N mg
N N
θ θ
θ θ
