Lecture 33 Solving Rigid-body Equilibrium Problems

Lecture 33: Solving rigid-‐body equilibrium problems

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Lecture Objective:

Apply conditions of equilibrium to rigid bodies

1st condition for equilibrium: Balanced Forces

2nd condition for equilibrium: Balanced Torques

Equilibrium of a rigid body

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1.) The rigid body must not be rotating and accelerating.2.) Draw the free body diagram. Rigid bodies cannot be represented as a point.3.) Choose the coordinate axis and a positive sense of rotation. 4.) Convert the forces into its component forms.5.) Note that for forces acting at one point, the torque at that point is zero.6.) Write equations expressing equilibrium position.

7.) Count the number of unknowns and equations.8.) Do the math.

Example 12.1 (Tipler)

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Given:length, 3 mboard mass, 2 kgbox mass, 6 kgx1, 2.5 mx2, 0.5 mtarget variable, F1 and F2

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Apply the two conditions of equilibrium

F1 + F2 - mboardg - mboxg = 0 First condition-F1L + mboardgL/2 + mboxgx2 + F2 x(0) = 0 Second condition

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can also be written as:F1 = 1/2mboardg + x2mbox/L

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Two equations with two unknowns so just substitute F1 to getF2 = mboard + mboxg - F1 = 1/2mboardg + mboxg -x2mboxg/L

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Two equations with two unknowns so just substitute F1 to getF2 = mboard + mboxg - F1 = 1/2mboardg + mboxg -x2mboxg/L

The scale readings will be:F1 = 19.6 NF2 = 58.9 N

Sample Problem Sir Lancelot is trying to rescue Lady Elayne from Castle Von Doom by climbing a uniform ladder that is 5.0m long, weighing 180N. Lancelot weighs 800N, stops a third of the way up the ladder. The bottom of the ladder rests on a horizontal stone ledge and leans across the moat in equilibrium against a vertical wall that is frictionless because of a thick layer of moss. The ladder make an angle of 53.1o with horizontal, conveniently forming a 3-‐4-‐5 right triangle.

(a) Find the normal force and frictional force on the ladder at its base. (b) Find the minimum μs needed to prevent slipping at the base. (c) Find the magnitude and direction of the contact force on the ladder at the

base.

FBD of Sir Lancelot and ladder

(a) Use the 1st condition for equilibrium: FBD of Lancelot and ladder

Use the 2nd condition for equilibrium:

The contact force at B is the superposition of the normal force and the static friction force

Seatwork

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SW 1: Clea’s weight SW2: x (cg)

Seatwork 3, 4 and 5:

A B

SW 3: Draw the FBD of the situation SW 4: Calculate the maximum weight you can place without breaking the cable. (use 1st condition) SW 5: Where should you put the weight? (use 2nd condition, calculate for CG)

Weight, w

Seatwork answers

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Apply the first and second conditions of equilibrium to Clea

The normal force at her feet is greater than at her rear feet, so her center of gravity is closer to her front feet ☺

Seatwork 1 and 2

A B

SW 3: Draw the FBD

wbar = 350N

TA = 500N TB = 400N

w = ???

Weight, w

SW4: What is the heaviest weight the you can put in the bar without breaking the cable? (add another weight)

ΣFy =0 -‐w -‐wbar + TA + TB= 0 -‐w – 350N + 500N + 400N = 0 w = 550N

SW 3: Draw the FBD

wbar = 350N

TA = 500N TB = 400N

w = ???

SW 3: Draw the FBD

wbar = 350N

TA = 500N TB = 400N

w = ???

SW5: where should you put it? We must put it at cg, solve for cg using the 2nd condition:Στ = 0

-‐TALcg + TB(1.5m-‐Lcg)–wbar(0.75m-‐Lcg) = 0 (-‐500N-‐400N+350N)Lcg = (0.75m)(350N)-‐(1.5m)(400N) Lcg = 0.614m

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Lecture 33 Solving Rigid-body Equilibrium Problems