Two-Body System: Two Hanging Masses

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Newton’s Laws and Pulley Systems D u lk u – P h ys ic s 2 0 – U n it 2 ( D yn am ic s) – T o p ic I Specific Outcome: Specific Outcome: Specific Outcome: Specific Outcome:

i. I can apply Newton’s laws of motion to solve, algebraically, linear motion problems in horizontal, vertical and inclined planes near the surface of Earth, ignoring air resistance.

Newton’s Laws and Pulley Systems

D u lk u – P h ys ic s 2 0 – U n it 2 ( D yn am ic s) – T o p ic I

Two-Body System: Two Hanging Masses Two-Body System: One

Hanging Mass Three-Body System: Two Hanging Masses

Tension in a String

We will deal with three different pulley systems

Each pulley system: 1) is frictionless, and

2) has a net force supplied only by the hanging mass or masses

Mass of the pulley itself is insignificant Two-Body System: Two Hanging Masses

This first system has 2 competing hanging masses

The system’s net force is the magnitude of the difference in both of their F_g values

The pulley turns in the direction of the heavier mass; both masses accelerate together as a system

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Steps to solve for this system:

1) Calculate F_netfrom the difference in F_g between the two hanging masses 2) Calculate magnitude of acceleration

from F_net and the system’s total mass 3) Directions: the larger mass is moving down, the smaller mass is moving up Two-Body System: Two Hanging Masses

Dulku – Physics 20 – Unit 2 (Dynamics) – Topic I

ex. Two masses are connected by a string over a light, frictionless pulley. Mass A is 55 kg, while mass B is 75 kg. Determine the acceleration of each mass.

Two-Body System: Two Hanging Masses

F_gA= m_Ag = (55 kg)(9.81 m/s2_{) = 539.55 N}

= 196.2 N

F_net= |F_gA- F_gB| = |539.55 N – 735.75 N| F_gB= m_Bg = (75 kg)(9.81 m/s2_{) = 735.75 N}

ex. Two masses are connected by a string over a light, frictionless pulley. Mass A is 55 kg, while mass B is 75 kg. Determine the acceleration of each mass.

Two-Body System: Two Hanging Masses

m_system= m_A+ m_B= 55 kg + 75 kg = 130 kg a_A= 1.5 m/s2_[up] _a B = 1.5 m/s2[down] a_system= 196.2 N 130 kg = 1.509…m/s 2 = F_net m_system

This second pulley system has one hanging mass that supplies all of the net force (F_g= F_net) The magnitude of acceleration

is the same for the surface mass and the hanging mass

Two-Body System: One Hanging Mass

m_A mB

The pulley turns in the direction of the hanging mass, which moves down

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The surface mass moves toward the pulley, ignoring friction

Steps:

1) Find F_netfrom F_gof the hanging mass 2) Find acceleration of the system from the

combined mass of both objects

Dulku – Physics 20 – Unit 2 (Dynamics) – Topic I ex. A 5.00 kg lab cart is accelerated on a

frictionless table by a 2.00 kg mass accelerating straight down, as shown. What is the acceleration of the lab cart? 5.00 kg 2.00 kg m_system= 5.00 kg + 2.00 kg = 7.00 kg F_net= F_g= mg = (2.00 kg)(9.81 m/s2_{) = 19.62 N} 19.62 N 7.00 kg a_system= Fnet m_system = = 2.80 m/s 2 a = 2.80 m/s2_[right]

ex. Find the acceleration of the two masses shown in the diagram. Assume there is no friction.

2.0 kg 1.0 kg m_system= 1.0 kg + 2.0 kg = 3.0 kg F_net= F_g= mg = (1.0 kg)(9.81 m/s2_{) = 9.81 N} 9.81 N 3.0 kg a_system= Fnet m_system = = 3.3 m/s 2 a_cart= 3.3 m/s2_[right] _a hanging= 3.3 m/s2[down]

ex. Find the acceleration of the two masses shown in the diagram. The cart experiences 5.0 N of friction.

10 kg 2.0 kg m_system= 2.0 kg + 10 kg = 12 kg F_g= mg = (2.0 kg)(9.81 m/s2_{) = 19.62N} 14.62 N 12 kg a_system= Fnet m_system = = 1.2 m/s 2 a_cart= 1.2 m/s2_[right] _a hanging= 1.2 m/s2[down] F_net= F_g+ F_f= 19.62N + (-5.0 N) = 14.62 N

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One last pulley system involves:

two competing hanging masses a surface mass that is pulled in the

direction of the heavier hanging mass all 3 masses accelerate together (system)

Three-Body System: Two Hanging Masses

mA mB

mC

Steps to solve a two pulley system:

1) Calculate F_netfrom the difference in F_g between the two hanging masses

2) Calculate acceleration from F_netand the combined mass of the system

3) Assign directions based on which hanging mass is heavier

Dulku – Physics 20 – Unit 2 (Dynamics) – Topic I ex. A 15 kg mass lies on a surface. The mass is

connected to a 10 kg mass and a 13 kg mass, each hanging by pulleys on each end. Calculate the magnitude of acceleration of the pulley system.

F_gA= m_Ag = (10 kg)(9.81 m/s2_{) = 98.1 N}

= 29.43 N

F_net= |F_gA- F_gB| = |98.1 N – 127.53 N| F_gB= m_Bg = (13 kg)(9.81 m/s2_{) = 127.53 N}

Dulku – Physics 20 – Unit 2 (Dynamics) – Topic I ex. A 15 kg mass lies on a surface. The mass is

connected to a 10 kg mass and a 13 kg mass, each hanging by pulleys on each end. Calculate the magnitude of acceleration of the pulley system.

m_system = 15 kg + 10 kg + 13 kg = 38 kg = 0.77 m/s2 a_system= 29.43 N 38 kg = F_net m_system

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Tension in a pulley string is applied force After finding acceleration, determine tension

by doing the following:

1) Choose one of the hanging masses of the system (and use its value of mass)

Tension in a String

2) Use F_net= F_A+ F_g(or ma = F_A+ mg) to

find tension, F_A = (75 kg)(-1.509…m/s2_{) – (75 kg)(-9.81 m/s}2₎ Tension in a String

F_net= F_A+ F_g becomes _{ma = F}

A+ mg if you chose the 75 kg mass, moving down:

F_A= ma – mg

= 6.2 x 102_{N [up]}

ex. Two masses are connected by a string over a light, frictionless pulley. Mass A is 55 kg, while mass B is 75 kg. Determine the tension in the string. [Refer to the first example in the whole lesson]

Tension in a String

ex. Two masses are connected by a string over a light, frictionless pulley. Mass A is 55 kg, while mass B is 75 kg. Determine the tension in the string. [Refer to the first example in the whole lesson]

= (55 kg)(1.509…m/s2_{) – (55 kg)(-9.81 m/s}2₎

if you chose the 55 kg mass, moving up:

F_A= ma – mg