ENG1040 Lec06

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Faculty of Engineering ENG1040 – Engineering Dynamics ENG1040 Engineering Dynamics Pulley Systems, Free Body Diagrams : Example Questions Dr Lau Ee Von – Sunway Lecture 6

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Transcript of ENG1040 Lec06

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Faculty of Engineering

ENG1040 – Engineering Dynamics

ENG1040Engineering Dynamics

Pulley Systems,Free Body Diagrams : Example Questions

Dr Lau Ee Von – Sunway

Lecture 6

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Past exam question

• Question 2, Sem 2, 2007• Draw free body diagrams for Blocks A and B

when Block B is translating and accelerating downwards.

• How is acceleration of Block B related to the acceleration of block A?

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Past exam question

• How do you approach a system with several pulleys (pulley system)?

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Lecture Outline

• Pulley systems:• How to gain mechanical leverage• Example Questions: FBD

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Pulley systems

• Pulley systems have been used for Millennia to reduce the force required to lift weights.

• Employed largely in sailing, they are believed to have been invented by Archimedes (200BC).

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Pulley systems

• Pulleys in everyday life:

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The Simplest Pulley system

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The simplest type of pulley system is shown here.

A free-body diagram of this pulley system shows that the total load is split into half on either pulley rope to maintain equilibrium.

But there must be a trade-off...

... The amount of work applied does not change.

dFEnergy

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Kinematics of Pulleys

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Therefore to raise the mass a distance d, the rope must be hoisted a distance 2d.

This also implies that if the rope is pulled with a velocity v, then the mass will move with a velocity v/2.

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Further improvements

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Often we want to pull down to pull a weight up – Gun tackle system

In this case, the beam has to support 1½ times the weight just to maintain equilibrium.

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We can make further improvements!

The Luff Tackle (shown here) has a mechanical advantage of 3.

Note, to maintain equilibrium, the tension in the rope is the same at all locations.

Pulley systems

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Once again, we can change the system so that we are pulling downwards to lift the weight.

Pulley systems

The more pulleys, the greater the mechanical advantage.Why stop at 4:1?

The greater the mechanical advantage, the further you have to pull the rope in order to shift the mass.

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Pulley systems

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• Note: the analysis described on the previous slides assumes that the pulleys are massless...

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Analysis procedure

1. Establish a coordinate system

2. Draw Free Body Diagram(s)• Graphical representation of all forces

acting on the system.

3. Establish known & unknown quantities

4. Apply Equation(s) of Motion in each direction

5. Evaluate kinematics to solve problem

Kinetics/Kinematics problems...

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B

A

Free body diagrams – Pulley system

Draw the FBD for the following pulley systems, assuming the pulleys and ropes are massless

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Free body diagrams – Pulley system

Draw the FBD for the following pulley systems, assuming the pulleys and ropes are massless

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Free body diagrams – Pulley system

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A

  

 

 

 

F

BC

   

   

  

 

Rope 1Rope 2

Question 2, Sem 1, 2012

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Question 3, Sem 2, 2011

Free body diagrams – Pulley system

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Kinematics

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𝑥𝐴+𝑥𝐵+𝑥𝐶=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

∆ 𝑠𝐴+∆ 𝑠𝐵+∆𝑠𝐶=0

𝑑𝑣𝐴

𝑑𝑡+𝑑𝑣𝐵

𝑑𝑡+𝑑𝑣𝐶

𝑑𝑡=0

Position vector from origin (fixed point)

Displacement = xfinal - xinitial

𝑎𝐴+𝑎𝐵+𝑎𝐶=0Equation for the acceleration relationship between masses

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Example Question

• Question 12.12 [Kinetics] (MECHANICS FOR ENGINEERS: DYNAMICS by Ferdinand P. Beer)

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The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between the blocks and the incline, determine

a) The acceleration of each blockb) The tension in the cable

x

y

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Example Question

• Question 12.12 [Kinetics]

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The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between the blocks and the incline, determine

a) The acceleration of each blockb) The tension in the cable

Ax

y

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Example Question

• Question 12.12 [Kinetics]

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The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between the blocks and the incline, determine

a) The acceleration of each blockb) The tension in the cable

A

)(30sin xAAo

Ax amgmTF

T

NFgmA

x

y

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Example Question

• Question 12.12 [Kinetics]

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The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between the blocks and the incline, determine

a) The acceleration of each blockb) The tension in the cable

Bx

y

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Example Question

• Question 12.12 [Kinetics]

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The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between the blocks and the incline, determine

a) The acceleration of each blockb) The tension in the cable

Bx

y

)(30sin3 xBBo

Bx amgmTF

T3

NFgmB

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Example Question

• Question 12.12 [Kinetics]

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The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between the blocks and the incline, determine

a) The acceleration of each blockb) The tension in the cable

)(30sin3 xBBo

B amgmT

)(30sin xAAo

A amgmT

How many unknowns do I have?

Do I have enough equations?

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Example Question

• Question 12.12 [Kinetics]

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The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between the blocks and the incline, determine

a) The acceleration of each blockb) The tension in the cable

If we consider the kinematics of the problem we can relate the acceleration of block A with the acceleration of block B:

03 )()( xBxA aa

constant3 BA xx

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Example Question

• Question 12.12 [Kinetics]

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The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between the blocks and the incline, determine

a) The acceleration of each blockb) The tension in the cable

)(30sin3 xBBo

B amgmT

)(30sin xAAo

A amgmT

)()( 3 xBxA aa aA = -3.30 m/s2

aB = 1.10 m/s2

T = 16 N

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Example Question

• Question 12.32 [kinetics]

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The weight of blocks A, B, and C are wa = wc = 20 lb, and wb=10 lb. Knowing that P = 50 lb and neglecting the masses of the pulleys and the effect of friction, determinea) The acceleration of each blockb) The tension in the cable.

1st step: Convert to SI units (see back of text book)

1 lb of force = 4.448 N of force

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Example Question

• Question 12.32 [kinetics]

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The weight of blocks A, B, and C are wa = wc = 20 lb, and wb=10 lb. Knowing that P = 50 lb and neglecting the masses of the pulleys and the effect of friction, determinea) The acceleration of each blockb) The tension in the cable.

mAg = mCg = 88.96 N mA = mC = 9.07 kg

mBg = 44.48 N mB = 4.54 kg

P = 222.4 N

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Example Question

• Question 12.32 [kinetics]

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The weight of blocks A, B, and C are wa = wc = 20 lb, and wb=10 lb. Knowing that P = 50 lb and neglecting the masses of the pulleys and the effect of friction, determinea) The acceleration of each blockb) The tension in the cable.

xy

A

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Example Question

• Question 12.32 [kinetics]

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The weight of blocks A, B, and C are wa = wc = 20 lb, and wb=10 lb. Knowing that P = 50 lb and neglecting the masses of the pulleys and the effect of friction, determinea) The acceleration of each blockb) The tension in the cable.

A

)(3 xAAx amTF x

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Example Question

• Question 12.32 [kinetics]

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The weight of blocks A, B, and C are wa = wc = 20 lb, and wb=10 lb. Knowing that P = 50 lb and neglecting the masses of the pulleys and the effect of friction, determinea) The acceleration of each blockb) The tension in the cable.

B

)(2 xBBx amTF x

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Example Question

• Question 12.32 [kinetics]

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The weight of blocks A, B, and C are wa = wc = 20 lb, and wb=10 lb. Knowing that P = 50 lb and neglecting the masses of the pulleys and the effect of friction, determinea) The acceleration of each blockb) The tension in the cable.

)(4 xCCx amTPF xC

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Example Question

• Question 12.32 [kinetics]

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The weight of blocks A, B, and C are wa = wc = 20 lb, and wb=10 lb. Knowing that P = 50 lb and neglecting the masses of the pulleys and the effect of friction, determinea) The acceleration of each blockb) The tension in the cable.

)(4 xCCamTP

)(2 xBBamT )(3 xAAamT

How many unknowns do I have?

Do I have enough equations?

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Example Question

• Question 12.32 [kinetics]

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The weight of blocks A, B, and C are wa = wc = 20 lb, and wb=10 lb. Knowing that P = 50 lb and neglecting the masses of the pulleys and the effect of friction, determinea) The acceleration of each blockb) The tension in the cable.

If we consider the kinematics of the problem we can relate the three acceleration terms.

First, we note that the pulley system is attached to the ground at this point.

We will measure the length of rope from this point.

0

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• Question 12.32 [kinetics]

We notice that two lengths of rope connect mass B to the fixed point.

Therefore, part of the rope’s length is defined as:

This is two times the distance from mass B to the fixed point.

Example Question

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The weight of blocks A, B, and C are wa = wc = 20 lb, and wb=10 lb. Knowing that P = 50 lb and neglecting the masses of the pulleys and the effect of friction, determinea) The acceleration of each blockb) The tension in the cable.

If we consider the kinematics of the problem we can relate the three acceleration terms.

bx20

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Example Question

• Question 12.32 [kinetics]

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The weight of blocks A, B, and C are wa = wc = 20 lb, and wb=10 lb. Knowing that P = 50 lb and neglecting the masses of the pulleys and the effect of friction, determinea) The acceleration of each blockb) The tension in the cable.

If we consider the kinematics of the problem we can relate the three acceleration terms.

We notice that three lengths of rope connect mass A to the fixed point.

Therefore, part of the rope’s length is defined as:

This is three times the distance from mass A to the fixed point.

ax30

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Example Question

• Question 12.32 [kinetics]

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The weight of blocks A, B, and C are wa = wc = 20 lb, and wb=10 lb. Knowing that P = 50 lb and neglecting the masses of the pulleys and the effect of friction, determinea) The acceleration of each blockb) The tension in the cable.

If we consider the kinematics of the problem we can relate the three acceleration terms. Finally, we notice that four

lengths of rope connect mass C to the fixed point.

Therefore, part of the rope’s length is defined as:

This is four times the distance from mass C to the fixed point.

cx4

0

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Example Question

• Question 12.32 [kinetics]

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The weight of blocks A, B, and C are wa = wc = 20 lb, and wb=10 lb. Knowing that P = 50 lb and neglecting the masses of the pulleys and the effect of friction, determinea) The acceleration of each blockb) The tension in the cable.I can then sum all these lengths of rope together to form:

From this equation, I can determine an equation for velocity and acceleration...

constant423 cba xxx

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Example Question

• Question 12.32 [kinetics]

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The weight of blocks A, B, and C are wa = wc = 20 lb, and wb=10 lb. Knowing that P = 50 lb and neglecting the masses of the pulleys and the effect of friction, determinea) The acceleration of each blockb) The tension in the cable.First the velocity:

0423 dt

dx

dt

dx

dt

dx

dt

dx cba

0423 cba vvvv

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Example Question

• Question 12.32 [kinetics]

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The weight of blocks A, B, and C are wa = wc = 20 lb, and wb=10 lb. Knowing that P = 50 lb and neglecting the masses of the pulleys and the effect of friction, determinea) The acceleration of each blockb) The tension in the cable.Then the acceleration:

0423 dt

dv

dt

dv

dt

dv

dt

dv cba

0423 cba aaaa

I now have an equation relating the acceleration of the three weights.

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Example Question

• Question 12.32 [kinetics]

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The weight of blocks A, B, and C are wa = wc = 20 lb, and wb=10 lb. Knowing that P = 50 lb and neglecting the masses of the pulleys and the effect of friction, determinea) The acceleration of each blockb) The tension in the cable.

)(4 xCCamTP

)(2 xBBamT )(3 xAAamT

0423 cba aaa

Using these equations, I can solve the problem. Note that I have four equations and four unknowns.

aA = 8.9 m/s2

aB = 11.9 m/s2

aC = 12.6 m/s2

T = 27 N