Abeka Physics Section 9.2 Flashcards
Consider this situation: a ball resting on a hillside cannot move horizontally into the hill no matter how hard you kick or hit the ball in that direction. Nevertheless, in response to a horizontal force, the ball will move: it will scoot along an inclined path toward the top of the hill. How would you predict the acceleration in this case?
To predict the amount of acceleration in this case, you resolve the applied force into components, one parallel to the hillside (in the direction of motion) and the other in the perpendicular direction (into the hillside). Since the ball cannot move into the hillside (the perpendicular motion), the force up the hill (parallel to the hillside) governs acceleration.
Describe how friction is the one element related to both horizontal and vertical forces
Friction is the one element related to both horizontal and vertical forces, and it does so by means of the normal force (Fn), which is the force a surface exerts perpendicularly against an object’s weight. The friction (Ft) between an object and the surface it rests depending on whether the object is moving or still: Ft = u x Fn
The relationship between friction force (horizontal) and the normal force (vertical) is a relationship between their magnitudes, not their directions
In other words, the sign of the normal force does not determine the sign of the friction force, nor vice versa.
Imagine pushing your dresser across a wooden floor. Name all the forces acting on the dresser.
The forces acting on the dresser are the force “F” that you apply to the dresser, the friction “Ft” between the dresser and the floor, the weight “Fw” of the dresser pressing down against the floor, and the normal force “Fn” of the floor pushing up against the dresser’s weight.
Describe the uses of Newton’s second law of motion, F = ma
> to solve motion problems involving one force acting on an object
to multiply forces acting on a single object (but instead of using F = ma, use Sigma(F) = ma, where Sigma (F) is the sum of all the forces acting on an object in a given axis
horizontal vectors cannot be combined with vertical vectors into the same equation in free-body diagrams
What are the horizontal forces acting when you are pushing your dresser?
> F: the force acting on the dresser
>Ft: the friction between the dresser and the floor
Show how you would use F=ma for the horizontal forces acting when you are pushing your dresser.
Sigma F”x” = F - Ft
(the sum of the horizontal forces equals the force of you pushing your dresser minus the friction force between your dresser and the floor)
>the reason for subtracting the friction force from force is that friction force acts in the opposite direction as the force
>the subscript “x” of Sigma F indicates only the forces in the horizontal direction
>to use Newton’s second law, set the sum of forces equal to “ma”
Sigma Fx = F - Ft = ma”x”
or simply
F - Ft = ma”x”
where “m” is the mass of the dresser and “a (subscript “x”) is the dresser’s horizontal acceleration
Apply Newton’s second law (F=ma) to vertical forces acting when you push your dresser across a wooden floor
Sigma F”y” = Fn - Fw = ma”y”
where m is the mass of the dresser
a (subscript “y”) is the dresser’s vertical acceleration
Fn is the normal force pushing from the floor up to the dresser
Fw is the weight of the dresser
>because the dresser is not accelerating up off of the floor, a (subscript “y”) = 0. Thus,
Fn - Fw = 0 so,
Fn = Fw (although they act in separate directions, their magnitudes are the same)
The horizontal forces and vertical forces are independent of each other except for the relationship between friction and the normal force: Ff = u x Fn. Since Fn = Fw and Fw = umg
Show and explain how this can be simplified.
This can be substituted into the horizontal forces equation above: F - umg = ma”x”
This is the general method for solving force problems. If the object is moving, u will be the coefficient of kinetic friction (u with subscript “k”); if the object is stationary, u will be the coefficient of static friction (u with subscript “s”)
A 35.0-kg dresser is being pushed with a horizontal force of 60.0 N. If the coefficient of kinetic friction is 0.145 between the floor and the dresser, find the dresser’s acceleration.
Set up an equation to relate the horizontal forces and an equation to relate the vertical forces:
Sigma F”x” = F - F”kf” = ma”x”
Sigma F”y” = F”n” - F”w” = ma”y”
Once again, since the dresser is not moving vertically, a”y” = 0, and this simplifies the vertical equation to
F”n” - F”w” = 0
F”n” = F”w”
Now, the two equations are
x-axis: F - F”fk” = ma”x”
y-axis: F”n” = F”w”
The values of f”kf”, F”n” and F”w” are unknown; however, known equations include F”w” = mg and F”kf” = u”k”F”‘n”. Since F”n” = F”w”, then F”kf” = u”k”mg. Thus,
F - u”k”mg = ma”x”
Now, substitute known values and solve. These are the known values: F = 60.0 N, m = 35.0 kg, and u”k” = 0.145. Thus,
60.0 N -(0.145)(35.0 kg)(9.80 m/s^2) = (35.0 kg)a”x”
a”x” = 0.293 m/s^2
Describe angled force
> if forces are applied at an angle to an object on a horizontal surface, these forces will act to change the normal force
for instance, if dresser is pushed with force at 30 degree angle below horizontal, some of applied force acts to push dresser downward into floor, which is same direction as weight acts
according to 3rd law of motion, floor exerts upward reaction force that is equal to sum of both these forces
net result is that normal force (F”n”) is larger than if dresser were pushed with horizontal force
on other hand, if dresser is pulled with force angled at 30 degrees above horizontal, some of applied force pulls dresser upward, which is in opposite direction of weight. This decreases normal force exerted by floor on dresser.
to find vertical component of applied force, use trigonometric functions as shown in example below.
Imagine a block poised at the top of an inclined plane. Give all the physics characteristics.
> block is unstable because ramp does not support all of block’s weight; instead, some of block’s weight acts to make block slide down incline
direction of possible motion is not same as direction of weight vector
to find net force in direction of motion, forces must be considered parallel and perpendicular to incline
parallel component, F”w||”, pulls block against surface and determines resistive force of friction
since incline does not support all of block’s weight, normal force (F”n”) will be lower than if block sat on flat surface
this explains why box is easier to push down ramp than across horizontal floor
since F”w” is perpendicular to ground, angle between F”w” and ground is 90 degrees
since angles of any triangle must sum to 180 degrees, theta and alpha must sum to 90 degrees
since F”w||” is parallel to ramp face, and F”w_|” is perpendicular to ramp, this makes right triangle. thus sum of alpha and beta must also be 90 degrees
since alpha + theta = 90 degrees and alpha + beta = 90 degrees, it follows that alpha + theta = alpha + beta
theta = beta
this will be true for all incline. therefore, weight components can be described as follows: F”w||” = F”w” sin theta
F”w|_” = F”w” cos theta
