Lecture 10 Applying Newton's Laws

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LECTURE 10 APPLYING NEWTON’S LAWS 5.1 Equilibrium The life-support unit strapped to the back of astronaut Schmitt weighed 300 lbs. Static equilibrium During his training, a 50-lb mock-up was used 5.2 Dynamics and Newton’s second law with unforeseen consequences. 5.3 Mass and weight Apparent weight Weightlessness 5.4 Normal forces Learning objectives

nd ! Apply Newton’s 2 law to equilibrium problems to solve for direction and magnitude of unknown forces. nd ! Apply Newton’s 2 law in one and two dimensions.

! Distinguish mass, weight, and apparent weight. Quiz: 5.1-1 (Knight P5.5 modified)

! When you bend your knee, the quadriceps muscle is stretched. This increases the tension in the quadriceps tendon attached to your kneecap (patella), which, in turn, increases the tension in the patella tendon that attaches your kneecap to your lower leg bone (tibia). Simultaneously, the end of your upper leg bone (femur) pushes outward on the patella. The figure shows how these parts of a knee joint are arranged.

! Assuming that the kneecap is at rest, what is the magnitude of the net force on it in N? Quiz: 5.1-1 (Knight P5.5 modified) answer

! Assuming that the kneecap is at rest, what is the net force on it in N?

! An object in static equilibrium or in dynamic equilibrium satisfies

∑ "# = %&# = 0 and ∑ "( = %&( = 0

! Since the kneecap is in static equilibrium, the magnitude of the net force on it is 0 N. Discussion question: 5.1-1 (Knight P5.5 modified)

! Draw a free-body diagram of the kneecap, labeling the forces with our notation convention, type%&,(). You can ignore the weight of the kneecap. Discussion question: 5.1-1 (Knight P5.5 modified) answer

! Draw a free-body diagram of the kneecap, labeling the forces with our notation convention, type%&,(). You can ignore the weight of the kneecap.

*+,

./, *-, Quiz: 5.1-2 (Knight P5.5)

! What size force in N does the femur exert on the kneecap if the tendons are oriented as in the figure and the tension in each tendon is 60 N? You can ignore the weight of the kneecap. Quiz: 5.1-2 (Knight P5.5) answer

! What size force in N does the femur exert on the kneecap if the tendons are oriented as in the figure and the tension in each tendon is 60 N? You can ignore the weight of the kneecap.

! ∑ "# = 0

! &'( cos , + ./(, # = 0

! ./(, # = −&'( cos , = − 60 N cos 42° = −44.6 N

! ∑ "8 = 0

! &9( − &'( sin , + ./(, 8 = 0 B &9( ! ./(, 8 = &'( sin , − &9( = 60 N sin 42° − 60 N = −19.85 N A @ @ @ @ ! ./( = ./(, # + ./(, 8 = −44.6 N + −19.85 N = 49 N

! The femur force points 24° counterclockwise from the negative x-axis. , = 42° ./( &'( 5.2 Dynamics and Newton’s second law

! Newton’s second law can be decomposed into components.

∑ "# = %&# and ∑ "' = %&' 5.3 Mass and weight / apparent weight / weightlessness

! Weight is the gravitational force exerted on an object by a planet:

! = #$ '%(

! Apparent weight, or our sensation of the !%& gravitational force comes from forces that balance it.

! Weightlessness happens when an object is in free-fall where gravitational force is the only force acting on an object and its apparent weight is zero. Weight vs. mass

! Mass is an intrinsic property of an object, a measure of its inertia, and it does not matter where it is.

! Weight of an object (! = #$) depends on where it is. ! Its weight on the moon is about 1/6 of that on Earth. ! The life-support unit strapped to the back of astronaut Schmitt weighed 300 lbs on Earth and had a mass of 136 kg. ! During his training, a 50-lb mock-up with a mass of 23 kg was used. ! Although this effectively simulated the reduced weight of the unit on the moon, it did not correctly mimic the unchanged mass. ! It was more difficult to accelerate the 136-kg unit (by jumping or twisting suddenly) on the moon than it was to accelerate the 23-kg unit on Earth. Quiz: 5.3-1

! If you are on a bathroom scale in a descending elevator that is coming to a stop, which of the free-body diagrams (without labeling) is correct one of you?

A. B. C. D. E. Quiz: 5.3-1 answer

! If you are on a bathroom scale in a descending elevator that is coming to a stop, which of the free-body diagrams (without labeling) is correct one of you?

! The acceleration of the elevator and you in it is upward, so the net force on your is also upward.

! The normal force by the elevator floor on you must be larger in magnitude than the magnitude of the weight.

!"#

$%#

A B C D E Quiz: 5.3-2

! If you are on a bathroom scale in a descending elevator that is coming to a stop, is the scale measurement greater than, less than, or equal to your weight? A. Greater then B. Less than C. Equal to Quiz: 5.3-2 answer

! If you are on a bathroom scale in a descending elevator that is coming to a stop, is the scale measurement greater than, less than, or equal to your weight?

! The magnitude of the normal force on you by the scale is greater than the magnitude of your weight. !"# ! The scale is measuring the normal force that you exerts on the scale, which has the same magnitude as the normal force on you by the scale. &'#

! !"# = !#" > &'#

! So the apparent weight, !"# , is greater than your weight. Quiz: 5.3-3

! A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a bathroom scale. Sadly, the elevator cable breaks. What is the reading on the scale in N during the few seconds it takes the student to plunge to his doom? Quiz: 5.3-3 answer

! A 50-kg student (mg = 490 N) gets in a 1000- kg elevator at rest and stands on a bathroom scale. Sadly, the elevator cable breaks. What is the reading on the scale in N during the few seconds it takes the student to plunge to his doom?

! 0 N

! The student is free-falling, under the influence of the gravity alone. So his apparent weight is zero.

! FYI: Fortunately, elevators in the real world have so many safety features that this kind of stuff never happens. Group exercise: cylinder free body diagram

! A cylinder with mass ! = 1.350 kg is prevented from rolling down a ramp at 30.0° to the horizontal by a block of wood. You can ignore the friction on the cylinder by the ramp.

! Draw a free body diagram for the cylinder. 1.350 kg

30.0° Group exercise: cylinder free body diagram answer

! A cylinder with mass ! = 1.350 kg is prevented from rolling down a ramp at 30.0° to the horizontal by a block of wood. You can ignore the friction on the cylinder by the ramp.

! Draw a free body diagram for the cylinder. 1.350 kg

+,- +.- ⊗ )⃗ = 0 30.0°

/0- Quiz: 3.4-1

! A cylinder with mass ! = 1.350 kg is prevented from rolling down a ramp at 30.0° to the horizontal by a block of wood. You can ignore the friction on the cylinder by the ramp. 1.350 kg

! What is the magnitude of the force of the block on the cylinder?

! Enter your answer in N, but only include the number. 30.0° ! If you finish early, calculate the magnitude of the force of the ramp on the cylinder. Quiz: 3.4-1 answer

! A cylinder with mass ! = 1.350 kg is prevented from rolling down a ramp at 30.0° to the horizontal by a block of wood. You can ignore the 1.350 kg friction on the cylinder by the ramp.

! What is the magnitude of the force of the block on the cylinder?

! Pick a coordinate system. 30.0° ! Split ()* into + and , components.

! (-*,/ = −!1 sin 30° ? ! (-*,6 = −!1 cos 30° C* ?)* nd ∑ ;< ! Newton’s 2 law in the + direction is 9/ = = ⊗ ! Since 9/ = 0, therefore ∑ >/ = 0, so ?)* − (-*,/ = ?)* − !1 sin 30° = 0 (-*,/ (-*,6 ! Therefore ?)* = !1 sin 30° = 6.6 N , 30° + (-* Quiz: 3.4-2

! A cylinder with mass ! = 1.350 kg is prevented from rolling down a ramp at 30.0° to the horizontal by a block of wood. You can ignore the friction on the cylinder by the ramp. 1.350 kg

! What is the magnitude of the force of the ramp on the cylinder?

! Enter your answer in N, but only include the number. 30.0° Quiz: 3.4-2 answer / demo

! A cylinder with mass ! = 1.350 kg is prevented from rolling down a ramp at 30.0° to the horizontal by a block of wood. You can ignore the 1.350 kg friction on the cylinder by the ramp.

! What is the magnitude of the force of the block on the cylinder?

! Pick a coordinate system. 30.0° ! Split ()* into + and , components.

! (-*,/ = −!1 sin 30° ? ! (-*,6 = −!1 cos 30° @* ?)* ∑ ; nd < ! Newton’s 2 law in the + direction is 9 = 6 = ⊗ ! Since 96 = 0, therefore ∑ >6 = 0, so ?@* − (-*,6 = ?@* − !1 cos 30° = 0 (-*,/ (-*,6 ! Therefore ?@* = !1 cos 30° = 11.5 N , 30° + (-*