Equilibrium of Forces and Tor ques

Equilibrium of Forces

Example 1: A bowling pin in stable equilibrium (Translational Equilibrium):

∑ F⃑ y=0 ¿ ∑ F⃑x=0

F⃑N=F⃑G

∑ F⃑=0

Example 2: A mass is hung by two cables from the ceiling as shown. Calculate the force of tension in both cables.

Method 1: Solve for F⃑T1 and F⃑T2 using sine law.

sin 95 °mg

= sin 20 °F⃑T 2

=sin (180 °−95 °−20 ° )

F⃑T1

∴ F⃑T2=mg sin 20°

sin 95 °=

(5 kg ) (9.8 N /kg ) sin 20°sin 95 °

=17 N

∴ F⃑T1=mg sin 65°

sin 95 °=

(5 kg ) (9.8 N /kg ) sin 65°sin 95 °

=45 N

Method 2: “Components”

¿ x>¿∑ F⃑x=0 ∴ ´F⃑T1x=⃗⃑FT2 x ¿¿

⇒ F⃑T1 cos70 °=F⃑T2 cos25 ° ∴ F⃑T1=F⃑T 2cos 25°

cos70 °(1)

¿ y>¿∑ F⃑ y=0 ∴ F⃑T1 y+ F⃑T 2 y=F⃑G ( forceup ↑=↓ forcedown )

⇒ F⃑T1 sin70 °+ F⃑T2 sin 25°=5 g (2)

Substitute(1) into(2): ( F⃑T2 cos25 °cos70 ° )sin 70°+ F⃑T2 sin 25 °=5 g

⇒ F⃑T2 (cos25 ° tan 70°+sin 25 ° )=5 g

∴ F⃑T2=5 g

cos25 ° tan 70 °+sin 25 °=17 N , 25 ° abovethe horizontal alongCable 2

∴ F⃑T1=F⃑T 2cos 25°

cos70 °=(17 N ) cos25 °

cos70 °=45 N , 70° above thehorizontal alongCable1

Equilibrium of Torques

Torque=Force× Lever Arm ⇒ T=F ×d⊥ (Torque is measured∈N ∙m )

Example 1: Calculate T P.

a)

T P=F ×d⊥=−(5N ) ( 0.30 m)=−1.5 Nm (¿ ↷1.5 Nm)

Note: “coming out of the page” is “+”

To tell which

direction is “positive,” grab the z axis with your right hand. The direction in which your fingers curl is positive, as shown below.

b)

T P=F⊥×d=−(5N )sin 30 ° × (0.30 m )

¿−0.75 Nm (¿ ↷0.75 Nm)

¿ T P=F ×d⊥=− (5N )× (0.30 m )sin 30 °

¿−0.75 Nm (¿ ↷0.75 Nm)

c)

T P=0 Nm

since F is line with the pivot (d⊥=0).

Example 2: Where does Big Brother sit to balance the seesaw?

For rotational equilibrium,

∑T=0 ∴ ↷T P

=↶T P

⇒ FG1 ∙ d1=FG2 ∙ d2

⇒ (19 g ) (2.0 )=(84 g ) ( x ) ∴ ¿ the pivot , P ¿

Example 3: A diver stands on the diving board. Calculate the forces on supports A and B.

Solution

∑ F y=0 ∴ forcesup ↑=↓ forcesdown

FG1+FG2=F A+FB

⇒ 65 g+10 g=F A+FB

∴ 75 g=F A+FB (1)

∑T=0 ∴ ↷T P

=↶T P

Decide on the position of the pivot P first.

(Hint: Placing it on one of the unknown forces eliminates that force)

Pivot at A : FG1 d1+FG2d2=FB d B ⇒ (65 g ) (3.5 )+(10 g ) (1.75 )=FB (1.5 )

∴ FB=(65 g ) (3.5 )+(10 g ) (1.75 )

1.5=1600 N up (compression) (2)

Substitute (2 )into (1 ) : 75 g=F A+1600

∴ F A=75 g−1600=−870 N=870 N down(tension)

Note: You could also solve this problem by putting the pivot at B instead of at A.

Example 4: Arman places his lump of gold on the edge of the table as shown. Does the table flip over?

Table: 10 kg

Solution

∑T B=0 so ↷T B

=↶T B

⇒ FN A d A+FG1d1=FG2 d2

⇒ FN A (1.5 )+ (65 g ) ( 0.4 )= (10 g ) (0.75 )

∴ FN A=(10 g ) (0.75 )−(65 g ) (0.4 )

1.5=−120 N

∴ ¿

Where should Arman place his gold so that the table does not tip over?

Solution: At the point of tipping, FN A=0 N .

FN A d A+FG1d1=FG2 d2 ⇒ (0 ) (1.5 )+(65 g )d1=(10 g ) (0.75 )

∴ ¿ the pivot ,∨1.37 m¿ the centre of thetable ¿

Example 5: The U-Hill Sign

a) Calculate the force of tension in the cable.

b) Calculate F x and F y at the pivot, P.

c) Calculate the resultant force on the pin at pivot P.

Solution

a) The system is at equilibrium, so ∑T P=0 and ↷T P

=↶T P

∴ FGB d B+ FGS dS=FT y dT

b) The forces pulling horizontally and vertically at the pivot are both caused by the beam, so we need to

find FPx and FPy, respectively. Considering the entire system,

∑ F x=0 ∑ F y=0

F →=← F ⇒ FPx=FT x

⇒ FPx=FT cos30 °=(61 )(√32 )

∴ ¿¿

F ↑=↓ F ⇒ FT y=FGB+FGS−F Py

∴ FP y=FGB+FGS−FT y

⇒ FPy= (0.325 g )+ (2.0 g )−(61 sin 30 ° )

∴ FP y=7.5 N down

Note: You could also solve for FPy using torque, with the pivot at the point where the cable attaches to

the beam, 0.70 m from the wall (this eliminates the tension force from our equation).

∑T P=0 so ↷T P

=↶T P

⇒ FGS d 'S=FGB d '

B+FP y d 'P

…where d ' represents the distance from this new pivot point.

∴ FP y=FGS d '

S−FGB d 'B

d 'P

=(2.0 g ) (0.3 )−(0.325 g ) (0.2 )

(0.7 )=7.5 N down

c) We need to find the resultant force at P from FPx and FPy:

∑ FP=√ FPx2+FPy

2=√(52 )2+(7.5 )2=53 N

tanθ=F Py

FPx

∴ θ=tan−1( FP y

F Px)=tan−1( 7.5

52 )=8.1° belowthe horizontal

Example 6: The Ladder Problem

a) How high up the 4.0 m ladder can the painter safely go?

∑T P=0 so ↷T P

=↶T P

⇒ FGl⊥d l+FGP⊥d P=FNW⊥dNW

⇒ ( FGl cosθ ) d l+( FGPcos θ ) dP=( FNW sin θ ) dNW

⇒ (12 g cos60 ° ) (2.0 )+ (70 gcos60 ° ) dP=( FNW sin 60 ° ) (4 )

⇒ 12 g+(35 g ) dP=F NW (2√3 ) (1)

∑ F x=0 so F ←=→ F ⇒ FNW=FF=μ F NG

Since ∑ F y=0 FNG=FGl+P=12 g+70 g

⇒ FN W=(0.20 ) (82 g )=16.4 g (2)

Substitute(2) into(1): 12 g+ (35 g ) dP=(16.4 g ) (2√3 )

∴ d P=(16.4 g ) (2√3 )−12 g

35 g=1.3 mup the ladder

b) Calculate the resultant force acting on the ladder at the pivot, P.

We can see from the diagram to the left that there are two

forces acting at the pivot: FF and FNG.

Given FF=16.4 g ¿ FNG=82 g ∑ F P=√ FF2+FNG

2

∴ ∑ FP=√ (16.4 g )2+ (82 g )2=820 N

tanθ=F NG

FF

∴ θ=tan−1( FNG

FF)=tan−1( 82 g

16.4 g )=79 ° above the horizontal

Example 7: Pushing Things Over

Calculate the minimum force which needs to be applied to the top edge of the box to tip it over. Assume the box is of uniform mass. If the ground has a coefficient of friction of 0.50, does the box tip over without slipping?

We can “extend” the arrow for applied force and measure the perpendicular distance from

the axis of rotation (ie. pivot) to the force arrow (refer to FG and F A in the free-body diagram to the

right).

At the point of tipping,

↷T P

=↶T P

so FA d A=FG dG ∴ F A=FG dG

d A

=(3.0 g ) (0.10 )

0.40=7.4 N

FF=μ F N=μ FG=(0.50 ) (3.0 g )=15 N>7.4 N

The value we just calculated above is the limiting or maximum (static) friction force. As long as the applied force is less than this, which it is in this case, the box tips over without slipping

Example 8: Determine the angle which the rectangular prism below can be tilted to before it reaches the point of instability. Assume it does not slide.

At the point of instability, the centre of mass is directly above the pivot:

tanθ=0.51

=0.5 ∴ θ=tan−1 (0.5 )=27 °

Example 9: Determine the tension in the cable.

sin θ= 45

cosθ=35

tan θ=43

↷T P

=↶T P

so FGl⊥d l+FGB⊥d B=FT⊥dT ⇒ FGlsin θ d l+FGB sin θ dB=FT cosθ dT

(8 g )( 45 ) (5 )+(5 g )( 4

5 ) (2.5 )=FT ( 35 )(5 ) ∴ FT=

32 g+10 g3

=42 g3

=140 N

*Here is a shortcut method to solve the problem. It does not require any trigonometry.

As in Example 7, we can simply measure the perpendicular distance from the axis of rotation, or pivot, to the line or arrow along which

the force acts (refer to FT, FB

, and F l in the diagram to the left).

↷T P

=↶T P

so FGld l⊥+FGB d B⊥=FT dT⊥

∴ FT=FGl d l⊥+FGB dB⊥

dT⊥

=(8 g ) ( 4 )+(5 g ) (2 )

3

P P

CM

∴ FT=140 N along the cable

Centre of Mass

Centre of Mass of a Complex Shape

Assume the shape has a uniform thickness of t and a uniform density of ρ. If 1 m2=1 N , find x.

∑T P=T 1P+T 2P ∴ FG x=FG1 d1+FG2 d2

⇒ [ (3 ×1 )+( 4 ×1 ) ] x=(3 ×1 )( 32 )+ (4 ×1 )(3+1

2 ) ⇒ 7 x=92+14=18

12

∴ x=1812

÷ 7=3714

m≈ 2.64 m

Applications on Stability

1. How to pack a knap sack : Put heavier stuff near the top. This provides some “forward” torque to counteract the “backward” torque caused by the force of gravity pulling down on you and on the backpack. As a result, you do not need to lean as far forward to keep upright or strain your back as much as when the heavier stuff is near the bottom.

For the triangle,

|CMB|=13|AB|

2. Getting up from a chair : What must you do to get up more easily? Lean forward so your centre of mass is closer to/over the pivot (ie. your heels).

3. Balances : How are the small masses on the bathroom scale able to balance off the person’s weight? The pivot is close to you and far away from the small masses, so the torque is the same even though you weigh more than the masses.

4. Placing a heavy dish in the middle of a large circular dining table: How would you do it? How could you balance off the torque caused by the heavy dish? Stick one leg out behind you, like a ballerina; this causes torque in the opposite direction.

Fun Stuff: How fast must a car be going around a bend before there’s a danger of it rolling over? What would be the centripetal force acting on the car? What is the coefficient of friction of the road?

At the point of rolling over,

¿ y>¿ FN=FG=mg ¿ ¿ x>¿ FF=FC=m v2

r(assume FC<FFmax=μ FN)

↷T CM

= ↶T CM

so FF dF⊥=FN d N⊥ ⇒ (m v2

r )d F⊥= (m g ) d N⊥

∴ v=√ gr dN⊥

dF⊥

=√ (9.8 ) (10 ) (0.95 )0.75

=11 m/ s

FC=m v2

r=

(1500 ) (11)2

10=19 000 N

FFmax=μ FN=FC ∴ μ=FC

FN

= 19 000(1500 ) (9.8 )

=1.27