Abeka Physics Section 11.3 pg. 163 - 170 Flashcards
List the 6 devices that have traditionally been known in physics as simple machines
>lever >inclined plane >wedge >screw >pulley >wheel and axle
Describe the lever
> one of the simple machines in physics
effective simple machine
usually thought of as long, slender, inflexible object and assumed as perfectly rigid
to accomplish useful work, lever must be positioned so that is pivots freely on point called fulcrum
if force exerted at other end, lever is then able to move load, object resting on one end of lever
force applied to lever is called input force
F(sub i)
for applied to load by lever is called output force F(sub o)
distance through which input force acts is called input distance d(sub i)
distance through which output force acts is known as output distance d(sub o)
distance from fulcrum to point of application of input force is called input lever arm r(sub i)
distance from fulcrum to load is called output lever arm r(sub o)
work done on lever is called input work W(sub i)
work done on load is called output work
W(sub o)
Explain how to derive the law of levers
> according to definition of work, work done on lever [input work W(sub i)] is equal to input force times input distance [distance through which input force acts d(sub i)]: W(sub i) = F(sub i) x d(sub i)
similarly, work done on the load [output work W(sub o)] equals output force times output distance [distance through which output force acts d(sub o)]: W(sub o) = F(sub o) x d(sub o)
if no energy losses in lever itself, law of conservation of energy requires that input work (energy transferred from source of F(sub i) to lever) is equal to output work (energy transferred from lever to load):
F(sub i) x d(sub i) = F(sub o) x d(sub o)
dividing both sides by F(sub o) x d(sub i) gives
F(sub i)/F(sub o) = d(sub o) x d(sub i)
since distance each object moves is proportional to length of lever arms,
d(sub o)/ d(sub i) = r(sub o)/ r(sub i)
substituting left side of equation
F(sub i)/F(sub o) = d(sub o)/d(sub i)
for left side of equation
d(sub o)/d(sub i) = r(sub o)/r(sub i)
produces
F(sub i)/F(sub o) = r(sub o)/r(sub i)
then after clearing fractions,
F(sub i) x r(sub i) = F(sub o) x r(sub o)
this final equation is known as the law of levers
in words, law of levers states that the input force times the length of the input lever arm equals the output force times the length of the output lever arm
this means that if lever has long input arm and short output arm, output force will be greater than input force
Describe and define mechanical advantage
> mechanical advantage - the ability of a machine to multiply force
2 kinds of mechanical advantage: actual and ideal
actual mechanical advantage (AMA) of any machine is the ratio of the output force to the input force: AMA = F(sub o)/F(sub i)
ideal mechanical advantage (IMA) of any machine is the mechanical advantage without regard to energy losses due to friction, flexing of the lever, or other causes
IMA is expressed as ratio of input distance to output distance: IMA = d(sub i)/d(sub o)
inverting equation
d(sub o)/d(sub i) = r(sub o)/r(sub i) gives
d(sub i)/d(sub o) = r(sub i)/r(sub o)
thus, ratio of distances in equation
IMA = d(sub i)/d(sub o) can be replaces by ratio of lever arms: IMA = r(sub i)/r(sub o)
both AMA and IMA are dimensionless numbers
Solve this: Mike and Dave are balanced on a seesaw. Mike weighs 45lb. and is sitting 6.0ft from the fulcrum. Dave is sitting 4.0ft from the fulcrum. (a) What is Dave’s weight? (b) What is the IMA of the seesaw?
(a) Let Mike’s side be the input side, and let Dave’s weight be represented by F(sub o). Therefore,
F(sub i) is 45lb., r(sub i is 6.0ft, and r(sub o) is 4.0ft. Solving for F(sub o) in the law of levers gives
F(sub o) = F(sub i) x r(sub i)/ r(sub o)
Substitution gives
F(sub o) = (45lb.)(6ft)/4ft
F(sub o) = 68 lb; thus, Dave weighs about 68 lb.
(b) The IMA of the seesaw is given by equation
IMA = r(sub i)/r(sub o)
IMA = 6ft/4ft
IMA = 1.5
List all the classes of levers
>class 1 lever >class 2 lever >class 3 lever >levers are divided into 3 classes based on relative position of input force, fulcrum, and load
Describe class 1 lever
> if input force and load are located on opposite sides of fulcrum
example: seesaw
can multiply either force or speed and distance, depending on relative lengths of input and output arms
if input arm is longer, there is gain in force
if input arm is shorter, there is gain in speed and distance
Describe class 2 lever
> if input force and load are located on same side of fulcrum, lever belongs to class 2 or 3
if load is between fulcrum and input force
AMA of class 2 lever is always greater than 1
always increases force on load
Describe class 3 lever
> if input force and load are located on same side of fulcrum, lever belongs to class 2 or 3
it input force is between fulcrum and load
AMA of class 3 lever is always less than 1
always reduces force
often useful because, as it rotates, load moves faster and farther than does input
the loss in force on load is offset by gain in speed and distance
Describe the inclined plane
> commonly known as the ramp
allows heavy objects to be raised by sliding or rolling them along slope instead of by lifting them straight up
IMA of inclined plane is ration of input distance d(sub i) to output distance d(sub o):
IMA = d(sub i)/d(sub o)
in ideal case, where friction is neglected, input force is equal and opposite to component of load’s weight directed down along plane, or
F(sub w) sin theta, where output force is F(sub w).
in practical case, where friction cannot be ignored, input force must be increased by force of kinetic friction.
normal force is F(sub w) cos theta, so input force becomes F(sub i) = F(sub w) sin theta + u(sub k) x F(sub w) cos theta
since output force is still F(sub w), AMA is expressed as
AMA = F(sub w)/F(sub w)sin theta + u(sub k) x F(sub w) cos theta
after factoring denominator and canceling
F(sub w), equation becomes
AMA = 1/sin theta +u(sub k) cos theta
Solve this: A 3750-lb safe must be raised to the top of a 6.0-ft platform. A winch is available that is rated at 1200 lb maximum. (a) If the safe can be rolled on steel rollers of negligible friction, what length ramp must be supplied to move the safe? (b) If the safe slips off the rollers halfway up the ramp, what force will be required to slide if up the rest of the way, assuming the coefficient of friction is 0.25?
(a) Find the length of the ramp, d(sub i), by means of the equation: sin theta = d(sub o)/d(sub i) >which implies d(sub i) = d(sub o)/sin theta >since F(sub i) = F(sub w)sin theta, then sin theta = F(sub i)/F(sub w) where F(sub i) is 1200 lb and F(sub w) is 3750 lb. >substitute to obtain sin theta = 1200 lb/3750 lb = 0.32 >given that d(sub o) = 6.0ft, substitute now into the equation for d(sub i) to obtain d(sub i) = 6.0ft/0.32 = 18.75ft = 19ft
(b) To find the input force against friction, use equation F(sub i) = F(sub w)sin theta + u(sub k)F(sub w)cos theta
>since sin theta = 0.32, then theta = 18.6629…degrees, and cos theta = 0.947417…substitution yields
F(sub i) = (3750lb)(0.32) + (0.25)(3750lb)(0.947417…)
F(sub i) = 2088.20…lb = 2100 lb
Describe the wedge
> familiar examples: implement used for splitting logs into firewood and edge of knife or of any other cutting blade is also a wedge
although wedge resembles 2 inclined planes fixed back-to-back, input force is applied not along inclined surface but against midline of wedge
IMA of wedge is ratio of length to thckness, where length is equivalent to input distance d(sub i), and thickness to output distance d(sub o)
thus, wedge has same IMA as equation
IMA = d(sub i)/d(sub o)
wedge usually operates in high-friction environment
due to complex nature of forces involved, AMA of wedge is not calculated here
Describe screw
> examples: ordinary screws and bolts, complex locking mechanisms employed in firearms, and hatches of spacecraft
often used in combination with lever
example of combination of lever and screw: screw jack, device commonly used to lift automobiles during mechanical repairs
input force is applied to lever arm along circular path with circumference given by C = 2pi(r) where r is both radius of circle and length of lever arm, and C is both circumference and input distance
in 1 full turn of screw, load is advanced distance p, corresponding to width of single ridge and valley (this distance is known as pitch of screw)
IMA of screw jack or any other screw is given by IMA = 2pi(r)/p
frictional forces operating on screw are complex, but are usually great enough to prevent screw from turning in reverse direction when input force is removed
screw jack has such high IMA that it is very useful machine for moving heavy loads over short distances
Solve this: A screw jack has 24in. handle and pitch of 0.20in. The jack is rate at 25 tons. A force of 150lb. on the handle is required to lift the full rated load. What are (a) the IMA and (b) the AMA of the jack?
(a) The IMA of a screw jack is given by equation
IMA = 2pi(r)/p
IMA = 2pi(24in.)/0.20in
IMA = 750
(b) Use equation AMA = F(sub o)/F(sub i), recognizing that F(sub o) is 25 tons, or 50,000lb. and F(sub i) is 150 lb. The AMA is therefore as follows: AMA = F(sub o)/F(sub i)
AMA = 50,000 lb/150 lb
AMA = 330 lb.
Describe the pulley
> single fixed pulley not very exciting device, as it merely reverses direction of force
in order to raise load 1ft., input force must be exerted through total of 2ft.
thus, IMA of single movable pulley is 2, and input force, if friction neglected, is only half the load
when several pulleys are used in combination, device is called block and tackle, and IMA is equal to number of strands lifting load
in practical applications, maximum IMA of about 8 can be achieved before frictional forces begin to offset advantage of additional pulleys
block and tackle is used on large cranes, tow trucks, and in elevators
forces involved are great enough that fatal injuries can result from flying end of broken cable
