Lecture 8- Lecture notes that represent a concise and complete outline of the most important points and ideas, especially those considered most important by the professor. It clarifies ideas not ful PDF

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Lecture notes that represent a concise and complete outline of the most important points and ideas, especially those considered most important by the professor. It clarifies ideas not fully understood in the text or elaborate on material that the text mentions only briefly....

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Y. Abranyos (LaTex)

Chapter 9: Static Equilibrium • The Two Conditions for Equilibrium • Solving Statics Problems • Examples

1

Y. Abranyos (LaTex)

9-1 The Conditions for Equilibrium An object with forces acting on it, but that is not moving, is said to be in static equilibrium. An object can be moving and still be in equilibrium. There are two conditions which must be satisfied for an object to be in equilibrium.

First condition for equilibrium The first condition for equilibrium is that the forces along each coordinate axis add to zero. In other words, the net force Fnet = 0, that is it must have zero acceleration but it can still move with constant velocity. Suppose you are driving a car with constant velocity. You are in equilibrium even though you are moving since you are not accelerating. Therefore,we must have. X

Fi = 0 ⇒

i

X

Fix = 0,

i

X

Fiy = 0,

i

X

Fiz = 0.

i

(9 -1) Three forces are applied to a tree sapling, as shown in Fig. 9-46, to stabilize it. If FA = 385 N and FB = 475 N, find FC in magnitude and direction. Solution The tree is in equilibrium, therefore X

F=0⇒

X

Fx = 0,

X

Fy = 0

X

⇒ ⇒

Fx = FA + FB cos(105) + FC x = 0 FC x = −FA − FB cos(105) = −262.1 N X Fy = FB sin(105) + FC y = 0 FC y = −F B sin(105) = 528.4 N FC =

q

F 2 Cx + FC2 y = 528.4 N

θ = tan−1

FC y = 60.3◦ below the negative x-axis FC x 2

Y. Abranyos (LaTex) (9 -12) Find the tension in the two cords shown in the figure below. Neglect the mass of the cords, and assume that the angle θ = 33◦ and the mass m is 190 kg. Solution

F1 T

θ

FT 2

θ mg

m

X

F=0⇒

X

X

Fx = FT 2 − FT 1 cos(33) = 0, ⇒ FT 2 = FT 1 cos(33)

X

Fy = FT 1 sin(33) − mg = 0 ⇒ FT 1 sin(33) = mg ⇒ FT 1 =

Fx = 0,

X

Fy = 0 mg = 3418 N sin(33)

FT 2 = FT 1 cos(33) = 2867 N

(9 -13) Find the tension in the two wires supporting the traffic light shown in

y θ2

θ1

FT2 θ2

FT1 θ1

mg

m

Solution

3

x

Y. Abranyos (LaTex)

X

F=0⇒

X

X

Fx = FT 1 cos(37) − FT 1 cos(53) = 0 ⇒ FT 2 = FT 1

Fx = 0,

X

Fy = 0 cos(37) cos(53)

X

Fy = FT 1 sin(37) + FT 2 sin(53) − mg = 0 ⇒ FT 1 sin(37) + FT 1 cos(37) tan(53) = mg mg ⇒ FT 1 = = 195 N sin(37) + cos(37) tan(53 cos(37) = 258 N FT 2 = FT 1 cos(53)

Second condition for equilibrium The second condition of equilibrium is that there be no torque around any axis; the choice of axis is arbitrary. It is possible for the net force Fnet = 0 (satisfy the first condition for equilibrium) and not be in equilibrium. In the figure below, the net force is zero but the object is not in equilibrium since it has angular acceleration. F

Not in Equilibrium ( α is not zero) F

We must also have τnet =

X

any axis

τi = 0 ⇒

X

τcounter−clock−wise −

X

axis

axis

4

τclock−wise = 0

Y. Abranyos (LaTex)

9-2: Solving Statics Problems Solving Equilibrium problems: 1. Choose one object at a time, and make a free-body diagram showing all the forces on it and where they act. 2. Choose a coordinate system and resolve forces into components. 3. Write equilibrium equations for the forces. 4. Choose any axis perpendicular to the plane of the forces and write the torque equilibrium equation. A clever choice here can simplify the problem enormously. 5. Solve We use the convention counter-clock-wise (CCW) “+” and clock-wise (CW) “-”. Examples: 1) A uniform beam of mass 20 kg and length L = 10 m is supported at each end and a box of mass 10 kg is resting 7 m from one end. Find the magnitude of the supporting forces F1

F2 Free−body diagram for the beam

m = 10 kg P

mb = 20 kg 7m 10 m

X

mbg

mg

Fyi = F1 + F2 − mb g − mg = 0 (Condition 1)

i

X

0 = F1 + F2 − 196 − 98 τi = LF2 − 0.7Lmg − 0.5Lmb g = 0 (Condition 2)

P

0 = F2 − 68.6 − 98 = 0, → F2 = 166.6 N F1 = 196 + 98 − F2 = 127.4 N 5

Y. Abranyos (LaTex) 2) A diver of mass 60 kg is standing at the end of a 10 m long diving board of mass 40 kg as shown in the figure below. Calculate the forces F1 and F2 acting on the diving board. F2

Diver

Free−body diagram for the board P

8m

2m

F1

X

mb g

msg

Fyi = −F1 + F2 − mb g − md g = 0 (Condition 1)

i

0 = −F1 + F2 − 392 − 588 X

τi = 0.2LF2 − 0.5Lmb g − Lmd g = 0 (Condition 2)

P

0 = 0.2F2 − 196 − 588 = 0, → F2 = 3920 N F1 = F2 − 588 − 392 = 2940 N. 3) A sign of mass 25 kg is suspended from a uniform beam of mass 15 kg as shown in the figure below. Find the tension in the rope and the force acting on the hinge at P.

Fv

37o P

P

Free−body diagram for the beam

Fh

37 o

mbg

6

mg

Y. Abranyos (LaTex)

X

Fxi = Fh − FT cos(37) = 0 (Condition 1 x)

i

0 = Fh − 0.8FT X

Fyi = Fv + FT sin(37) − mb g − ms g = 0 (Condition 1 y)

i

X

0 = Fv + 0.6FT − 147 − 245 τi = LFT sin(37) − 0.5Lmb g − Lms g = 0 (Condition 2)

P

0 = 0.6FT − 245 − 73.5 = 0, → FT = 530.8 N 0 = Fv + 0.6FT − 147 − 245 → Fv = 73.52 N 0 = Fh − 0.8FT → Fh = 424.6 N. (9 -29) A door h = 2.30 m high and w = 1.30 m wide has a mass of m = 13.0 kg. A hinge d = 0.40 m from the top and another hinge 0.40 m from the bottom each support half the door’s weight. Assume that the center of gravity is at the geometrical center of the door, and determine the horizontal and vertical force components exerted by each hinge on the door. w

w

F Ay d F Ax

x h

h FBx FBy d

mg

Solution Calculate torques about the bottom hinge, with counterclockwise torques as positive. X mgw w τ = mg × − FAx × (h − 2d) = 0 ⇒ FAx = = 55.2 N 2 2(h − 2d) O X

Fxi = FAx + FBx = 0 ⇒ FBx = −FAx = −55.2 N

X

Fyi = FAy + FBy − mg = 0 ⇒ FAy = Fby =

i

i

7

mg = 63.7 N 3

Y. Abranyos (LaTex) 4) A ladder of mass 20 kg is leaning against a wall as shown in the diagram below. The coefficient of static friction between the floor and the ladder and, between the wall and the ladder are both µs . A person of mass 60 kg is standing 2 m up the ladder. What is the value of µs ? Ffrw

FNw

ladder person 8m

Free−body diagram for the ladder

FNf ml g mpg P

Ffrf

6m

X

Fxi = Ffrf − FN w = 0 (Condition 1 x)

X

Fyi = FN f + Ff rw − mp g − ml g = 0 (Condition 1 y)

i

i

X

τi = 8FN w + 6Ff rw − 3ml g − 1.2mp g = 0 (Condition 2)

P

T hese give the following Ffrf = FN w , ⇒ Ffrf = µs FN f ⇒ FN w 0 = FN f + µs FN w − 588 − 196 0 = 8FN w + 6µs FN w − 1.2 × 588 − 3 × 196 Combine the above three equations FN w + µs FN w 784 = µs 1293.6 = 8FN w + 6µs FN w Divide the above equations 8 + 6µs = 1.65 µs + 1/µs 4.35µ2s + 8µs − 1.65 = 0 ⇒ µs = 0.187 8...