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SPH4U: Practice Problems Today’s Agenda Run and Hide

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Review: Pegs & Pulleys Used to change the direction of forces Used to change the direction of forces An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: The tension is the same on both sides! An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: The tension is the same on both sides! Fi F 1 = -T i ideal peg or pulley Fj F 2 = T j FF | F 1 | = | F 2 | massless rope

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Problem: Accelerometer A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizontal road, and has an acceleration a in the x direction. The string makes an angle with respect to the vertical (y) axis. Solve for in terms of a and g. A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizontal road, and has an acceleration a in the x direction. The string makes an angle with respect to the vertical (y) axis. Solve for in terms of a and g. a i

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Accelerometer... Draw a free body diagram for the mass: Draw a free body diagram for the mass: What are all of the forces acting? What are all of the forces acting? m T T (string tension) g mg (gravitational force) i

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Accelerometer... Using components (recommended): Using components (recommended): i: F X = T X = T sin = ma i: F X = T X = T sin = ma j: F Y = T Y mg j: F Y = T Y mg = T cos mg = 0 = T cos mg = 0 T gmggmg m amaama ji TXTX TYTY

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Accelerometer... Using components : Using components : i: T sin = ma i: T sin = ma j: T cos - mg = 0 j: T cos - mg = 0 Eliminate T : Eliminate T : gmggmg m amaama T sin = ma T cos = mg TXTX TYTY ji T

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Accelerometer... Alternative solution using vectors (elegant but not as systematic): Alternative solution using vectors (elegant but not as systematic): Find the total vector force F NET : Find the total vector force F NET : T gmggmg F F TOT m T T (string tension) g mg (gravitational force)

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Accelerometer... Alternative solution using vectors (elegant but not as systematic): Alternative solution using vectors (elegant but not as systematic): Find the total vector force F NET : Find the total vector force F NET : Recall that F NET = ma: Recall that F NET = ma: So So amaama T gmggmg m T T (string tension) g mg (gravitational force)

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Accelerometer... Let’s put in some numbers: Let’s put in some numbers: Say the car goes from 0 to 60 mph in 10 seconds: Say the car goes from 0 to 60 mph in 10 seconds: 60 mph = 60 x 0.45 m/s = 27 m/s. 60 mph = 60 x 0.45 m/s = 27 m/s. Acceleration a = Δv/Δt = 2.7 m/s 2. Acceleration a = Δv/Δt = 2.7 m/s 2. So a/g = 2.7 / 9.8 = So a/g = 2.7 / 9.8 = = arctan (a/g) = 15.6 deg = arctan (a/g) = 15.6 deg a

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Understanding A person standing on a horizontal floor feels two forces: the downward pull of gravity and the upward supporting force from the floor. These two forces (A)Have equal magnitudes and form an action/reaction pair (B)Have equal magnitudes but do not form an action/reaction pair (C)Have unequal magnitudes and form an action/reaction pair (D)Have unequal magnitudes and do not form an action/reaction pair (E)None of the above Because the person is not accelerating, the net force they feel is zero. Therefore the magnitudes must be the same (opposite directions. These are not action/reaction forces because they act of the same object (the person). Action/Reaction pairs always act on different objects.

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Angles of an Inclined plane ma = mg sin mg N The triangles are similar, so the angles are the same!

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Problem: Inclined plane A block of mass m slides down a frictionless ramp that makes angle with respect to the horizontal. What is its acceleration a ? A block of mass m slides down a frictionless ramp that makes angle with respect to the horizontal. What is its acceleration a ? m a

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Inclined plane... Define convenient axes parallel and perpendicular to plane: Define convenient axes parallel and perpendicular to plane: Acceleration a is in x direction only. Acceleration a is in x direction only. m a i j

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Inclined plane... Consider x and y components separately: Consider x and y components separately: i: mg sin = ma. a = g sin i: mg sin = ma. a = g sin j: N - mg cos = 0. N = mg cos j: N - mg cos = 0. N = mg cos gmggmg N mg sin mg cos amaama i j

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Problem: Two Blocks Two blocks of masses m 1 and m 2 are placed in contact on a horizontal frictionless surface. If a force of magnitude F is applied to the box of mass m 1, what is the force on the block of mass m 2 ? Two blocks of masses m 1 and m 2 are placed in contact on a horizontal frictionless surface. If a force of magnitude F is applied to the box of mass m 1, what is the force on the block of mass m 2 ? m1m1m1m1 m2m2m2m2 F

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Problem: Two Blocks Realize that F = (m 1 + m 2 ) a : Realize that F = (m 1 + m 2 ) a : Draw FBD of block m 2 and apply F NET = ma: Draw FBD of block m 2 and apply F NET = ma: F 2,1 m 2 F 2,1 = m 2 a m 1 m 2 F / (m 1 + m 2 ) = a m2m2 2,1 m2m2 m1m1 F F l Substitute for a : m2m2m2m2 (m1 + m2)(m1 + m2) m2m2 F 2,1 F

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Problem: Tension and Angles A box is suspended from the ceiling by two ropes making an angle with the horizontal. What is the tension in each rope? A box is suspended from the ceiling by two ropes making an angle with the horizontal. What is the tension in each rope? m

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Problem: Tension and Angles Draw a FBD: Draw a FBD: T1T1 T2T2 mgmg l Since the box isn’t going anywhere, F x,NET = 0 and F y,NET = 0 T 1 sin T 2 sin T 2 cos T 1 cos ji F x,NET = T 1 cos - T 2 cos = 0 T 1 = T 2 2 sin mg T 1 = T 2 = F y,NET = T 1 sin + T 2 sin - mg = 0

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Problem: Motion in a Circle A boy ties a rock of mass m to the end of a string and twirls it in the vertical plane. The distance from his hand to the rock is R. The speed of the rock at the top of its trajectory is v. A boy ties a rock of mass m to the end of a string and twirls it in the vertical plane. The distance from his hand to the rock is R. The speed of the rock at the top of its trajectory is v. What is the tension T in the string at the top of the rock’s trajectory? What is the tension T in the string at the top of the rock’s trajectory? R v T

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Motion in a Circle... Draw a Free Body Diagram (pick y-direction to be down): Draw a Free Body Diagram (pick y-direction to be down): We will use F NET = ma (surprise) We will use F NET = ma (surprise) First find F NET in y direction: First find F NET in y direction: F NET = mg +T F NET = mg +T T gmggmg y

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Motion in a Circle... F NET = mg +T Acceleration in y direction: Acceleration in y direction: ma = mv 2 / R mg + T = mv 2 / R T = mv 2 / R - mg T = mv 2 / R - mg R T v gmggmg y F = ma

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Motion in a Circle... What is the minimum speed of the mass at the top of the trajectory such that the string does not go limp? What is the minimum speed of the mass at the top of the trajectory such that the string does not go limp? i.e. find v such that T = 0. i.e. find v such that T = 0. mv 2 / R = mg + T v 2 / R = g Notice that this does not depend on m. Notice that this does not depend on m. R gmggmg v T= 0

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Understanding Two-body dynamics In which case does block m experience a larger acceleration? In case (1) there is a 10 kg mass hanging from a rope. In case (2) a hand is providing a constant downward force of 98.1 N. In both cases the ropes and pulleys are massless. In which case does block m experience a larger acceleration? In case (1) there is a 10 kg mass hanging from a rope. In case (2) a hand is providing a constant downward force of 98.1 N. In both cases the ropes and pulleys are massless. (a) (b) (c) (a) Case (1) (b) Case (2) (c) same m 10kg aa m F = 98.1 N Case (1)Case (2)

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Solution Solution m 10kg a l Add (a) and (b): m W g = (m + m W )a (a) (b) T = ma (a) m W g -T = m W a (b) l For case (1) draw FBD and write F NET = ma for each block: m W =10kg

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Solution l The answer is (b) Case (2). In this case the block experiences a larger acceleration T = 98.1 N = ma l For case (2) m 10kg a Case (1) m a F = 98.1 N Case (2)

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Problem: Two strings & Two Masses on horizontal frictionless floor: m2m2 m1m1 T2T2 T1T1 Given T 1, m 1 and m 2, what are a and T 2 ? Given T 1, m 1 and m 2, what are a and T 2 ? T 1 - T 2 = m 1 a (a) T 2 = m 2 a (b) Add (a) + (b): T 1 = (m 1 + m 2 )a a Add (a) + (b): T 1 = (m 1 + m 2 )a a a i Plugging solution into (b):

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Understanding Two-body dynamics Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tension in each of the strings? Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tension in each of the strings? (a) T 1 > T 2 > T 3 (b) T 3 > T 2 > T 1 (c) T 1 = T 2 = T 3 T3T3 T2T2 T1T1 3m 2m m a

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Solution Draw free body diagrams!! Draw free body diagrams!! T3T3 3m T 3 = 3ma T3T3 T2T2 2m T 2 - T 3 = 2ma T 2 = 2ma +T 3 > T 3 T2T2 T1T1 m T 1 - T 2 = ma T 1 = ma + T 2 > T 2 T 1 > T 2 > T 3

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Alternative Solution Alternative Solution T3T3 T2T2 T1T1 3m 2m m a Consider T 1 to be pulling all the boxes T3T3 T2T2 T1T1 3m 2m m a T 2 is pulling only the boxes of mass 3m and 2m T3T3 T2T2 T1T1 3m 2m m a T 3 is pulling only the box of mass 3m T 1 > T 2 > T 3

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Problem: Rotating puck & weight. A mass m 1 slides in a circular path with speed v on a horizontal frictionless table. It is held at a radius R by a string threaded through a frictionless hole at the center of the table. At the other end of the string hangs a second mass m 2. A mass m 1 slides in a circular path with speed v on a horizontal frictionless table. It is held at a radius R by a string threaded through a frictionless hole at the center of the table. At the other end of the string hangs a second mass m 2. What is the tension (T) in the string? What is the tension (T) in the string? What is the speed (v) of the sliding mass? What is the speed (v) of the sliding mass? m1m1 m2m2 v R

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Problem: Rotating puck & weight... Draw FBD of hanging mass: Draw FBD of hanging mass: Since R is constant, a = 0. Since R is constant, a = 0. so T = m 2 g m2m2 m2gm2g T m1m1 m2m2 v R T

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Problem: Rotating puck & weight... Draw FBD of sliding mass: Draw FBD of sliding mass: m1m1 T = m 2 g m1gm1g N m1m1 m2m2 v R T Use F = T = m 1 a where a = v 2 / R m 2 g = m 1 v 2 / R T = m 2 g

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