UNIT 4
Phys 2002
Worksheet 1 Static Equilibrium
Worksheet 1 Static Equilibrium Answers
Worksheet 2 Mechanical Properties of Solids
Worksheet 2 Mechanical Properties of Solids Answers

 

 

Conditions for Static Equilibrium
Worksheet 1 - Questions and Answers on Static Equilibrium

ladder

1. An object of mass, M, is suspended between a block of mass 10kg and a wall by a string which makes and angle of 30°with the vertical. The coefficient of friction between the table top and the block is 0.5. What is the maximum weight that can be suspended?

Fig 1.
Answer  
ladder

At X the forces are:

T1 at 60° to the horizontal
T2L - horizoontal
W - vertical
W = Mg

ladder

Resolving T1 into vertical and horizontal components

Vertical component = T1 Sin 60°
Horizontal component = T2LCos 60°

ladder Forces on 10kg block at Y
R = Normal reaction
Weight of block = mg
Weight of block = 100N
For equilibrium R = 100N
T2R = Tension on the string attached to the block
Tension on the horizontal string is constant so the block experiences a force of T2R = -T2L.
Fig 1b.
 

(a)

 

Since X and Y are in equilibrium:
At Y
∑F = 0
∑Horizontal forces = 0
∑Vertical forces = 0
Normal reaction of table top = Weight of M
Normal reaction of table top (R) = 100N
Maximum friction (T2R) = μR
(T2R) = (0.5) (100)
(T2R) = (50N)--- Eq1
Tension on the horizontal string is constant so the block experiences a force of T2.

Supplementary Explanations:
Recall from Newton's First and Second Laws the conditions for static equilibrium:
— The linear momentum (momentum)of its center of mass is zero.

Vertical forces: (using up as positive)
At X

∑Vertical forces = 0

T1 Sin 60° + (-W) = 0
T1 Sin 60° = W
T1 Sin 60° = mg
m = (T1 Sin 60° / 10) --- Eq2

Recall that we can treat with horizontal and vertical forces independently as a force has no component at right angles to it so vertical forces can have no horizontal forces and vice-versa.




Horizontal forces: (using right as positive)
At X

∑Horizontal forces = 0
T2L + T1 Cos 60°= 0
(-50N) + T1 Cos 60°= 0
(T2 points to the left)
(-50N) =- T1 Cos 60°
(-50) =- 0.5 T1
T1 = 100N--- Eq3

Substituting for T1 in Eq2 from Eq3
m = [(100N Sin 60°/ 10)]
m = 8.6 kg

ladderAlternative solution:
Since there is no resultant force, the three vectors must add up to zero, so they form a closed triangle.

Tan 60° = W/T2L

W= [(T2L) X (Tan 60°)]

W = 86N

W = Mg
M - 8.6 kg

     
Worksheet 1 Static Equilibrium
Worksheet 1 Static Equilibrium Answers
Worksheet 2 Mechanical Properties of Solids
Worksheet 2 Mechanical Properties of Solids Answers
 
ladder

2. A leaf of mass 1 gram hangs on a rafter by a spider's thread as in fig 2b. The wind blows horizontally and the thread makes an angle of 20° with the vertical. Calculate (a) the tension on the thread and (b) the force of the wind on the leaf.

Fig 2a.
Answer  
ladder

Forces on the Leaf [Fig 2 (i)]

Let T = Tension on the web
F = Force of the wind
W = Weight of the leaf

Weight of leaf (W) = mg
W = (1 X 10-3 kg) (10ms2)
W = (1 X 10-2 N)

Resolving T into vertical and horizontal components [Fig 3b(ii)]
Vertical Component = T Cos 20°
Horizontal component = T Sin 20°

Fig 2b.
 

(a) ∑F = 0

∑Vertical forces = 0
T Cos 20° + (-W) = 0
T Cos 20° = - (-W)
T Cos 20° = (1 X 10-2 N)
---Eq1
T = (1 X 10-2 N) / Cos 20°
T = (1.1 X 10-2 N)

∑Horizontal forces = 0
(-T Sin 20°) + F = 0
---Eq2
F = T Sin 20°
F = (3.6 X 10-3 N)

Double check Answer.
T Sin 20° = (1 X 10-2 kg) ---Eq1
T Cos 20° = - F ---Eq2 rearranged.
Dividing Eq1 by Eq 2
Tan 20° = (1 X 10-2 kg) / (- F)
(- F) = (1 X 10-2 kg) / Tan 20°
F = -(2.7 X 10-2 N)

ladderAlternative solution:
Since there is no resultant force, the three vectors must add up to zero, so they form a closed triangle.

Cos 20° = W/T

T = W/Cos 20°

T = (0.01N)/Cos 20°

T = (1.1 X 10-2 N)

Tan 20° = F/W

F = Tan 20°/W

F = (Tan 20°)/(0.01N)

F = (3.6 X 10-3 N)

Double check:
(1.1 X 10-2 N)2 = (3.6 X 10-3 N)2 + (1.0 X 10-2 N)2

     
Worksheet 1 Static Equilibrium
Worksheet 1 Static Equilibrium Answers
Worksheet 2 Mechanical Properties of Solids
Worksheet 2 Mechanical Properties of Solids Answers
 
ladder

3. A flower pot of mass 8kg is hung from two wires as show in in fig 3a. Find the tension in each wire.

Fig 3a.
Answer  
ladder

Forces on the junction

T1 = Tension on the right wire
T2 = Tension on the left wire
W = Weight of the flower pot

Weight of flower pot (W) = mg
W = (8 kg) (10ms2)
W = (80 N)

 

ladder

Resolving T1 into vertical and horizontal components
Vertical Component = T1 Sin 40°
Horizontal component = T1 Cos 40°

Resolving T2 into vertical and horizontal components
Vertical Component = T21 Sin 60°
Horizontal component = T2 Cos 60°

Fig 3b.
 

(a) ∑F = 0
∑Horizontal forces = 0
T1 Cos 40° + [-(T2 Cos 60°)] = 0
T1 = -[-(T2 Cos 60°) / (Cos 40°)]
T1 = 0.65 T2 ---Eq2

∑Vertical forces = 0
T1 Sin 40° + T2 Sin 60° + (-W) = 0
T1 Sin 40° + T2 Sin 60° = -(-80N)
T1 Sin 40° + T2 Sin 60° = (80N)
T1 Sin 40° = (80N) - T2 Sin 60°
T1 = {[(80N) / (Sin 40°) - (T2 Sin 60°)] / (Sin 40°)}
T1 = 124N - 1.35 T2 ---Eq1

Substuting for T1 from Eq1 into Eq2
0.65 T2 = 124N - 1.35 T2
124N = 1.35 T2 + 0.65 T2
124N = 2 T2
T2 = 62 N
T1 = 41N


ladderAlternative solution:
Since there is no resultant force, the three vectors must add up to zero, so they form a closed triangle.

Using (W/Sin 100° ) = (T1/Sin 30°) = (T2/Sin 50°)


(80N/Sin 100° ) = (T1/Sin 30°) = (T2/Sin 50°)

T1 = 80 Sin 30/Sin 100°

T1 = 41N

T2 = 80 Sin 50°/Sin 100°

T2 = 62N

     
Worksheet 1 Static Equilibrium
Worksheet 1 Static Equilibrium Answers
Worksheet 2 Mechanical Properties of Solids
Worksheet 2 Mechanical Properties of Solids Answers
 
ladder

4. Several men with a combined strength of 4000N pull horizontally on a girder of mass 5000kg suspended from a crane. Assuming that the rope stays horizontal during the operation, What is the maximum angle, θ, that the bean can be shifted?

Fig 4a.
Answer  
ladder

Forces on the beam

T1 = Tension on the rope
T1 = 4000N
T2 = Tension on the cable from the crane
W = Weight of the beam

Weight of beam (W) = mg
W = (5000 kg) (10ms2)
W = (50000 N)

ladder

Resolving T2 into vertical and horizontal components
Vertical Component = T2 Cos θ
Horizontal component = T2 Sin θ

Fig 4b.
 

(a) ∑F = 0
∑Horizontal forces = 0
T1 + [-(T2 Sin θ)] = 0
4000N = T2 Sin θ
Sin θ = 4000N / T2 ---Eq1

∑Vertical forces = 0
T2 Cos θ + (-W) = 0
T2 Cos θ = W
T2 Cos θ = 50000N
Cos θ = 50000N / T2 ---Eq2

Dividing Eq1 and Eq2
Sin θ = 4000N / T2
Cos θ = 50000N / T2
Tan θ = (4000N / T2) / (50000N / T2)
Tan θ = (4000N) / (50000N)
θ = 4.6°

Alternative Solution
Sin2 θ = (4000N)2 / T22) ---Eq4
Cos2 θ = (50000N)2 / T22) ---Eq4
Adding Eq1 and Eq2

Cos2 θ + Sin2 θ = (4000N)2 / T22) +(50000N)2 / T22)
1 = [(4000N)2) + (50000N)2]/ T22
T22 = [(4000N)2) + (50000N)2]
T2 = ± 50160N

From Eq2
Cos θ = 50000N / 50160N
θ = 4.6°
Doublecheck
From Eq1
Sin θ = 4000N / 50160N
θ = 4.6°

ladderAlternative solution:
Since there is no resultant force, the three vectors must add up to zero, so they form a closed triangle.

Tan θ = (T1/W)

Tan θ = (4000N/50000N)

θ = 4.6°

     
Worksheet 1 Static Equilibrium
Worksheet 1 Static Equilibrium Answers
Worksheet 2 Mechanical Properties of Solids
Worksheet 2 Mechanical Properties of Solids Answers
 
ladder

5. A 3m ladder having a uniform density and a mass 30kg rests against a frictionless vertical wall at an angle of 30°. The lower end rests on a flat surface where the coefficient of static friction is µ = 0.40. A painter with a mass M = 60kg is at the middle of the ladder. At what point in his journey up the ladder will it begin to slip?

Fig 5.1.
Answer  
ladder

Forces on the ladder

Fw = Normal reaction of the wall
WL = Weight of ladder acts at P (centre of the ladder)
Wm = Weight of man
Fn = Normal reaction of the ground
Ff = Frictional force on the ladder from the ground

Weight of ladder (WL) = mg
WL = (30 kg) (10ms2)
WL = (300 N)
Weight of man (Wm) = mg
Wm = (60 kg) (10ms2)
Wm = (600 N)

z = Length of ladder X Sin 30°
z = 3m Sin 30°
y = (Length of ladder / 2) X Sin 30°
y = 1.5m Sin 30°


Στ = 0
Taking Moments about A
[- (Fn X 0m) + (Fr X 0m) + [-(Wm X xm)] + [-(Wl X ym)] + (Fw X zm) = 0
[-(600N X xm)] + [-(300N X1.5m Sin 30°m)] + (Fw X 6m Cos 30°m) = 0
Fw ={[(600N X xm) + (300N X1.5m Sin 30°m)] / (3m Cos 30°m)}
Fw ={[(600x) + (225N)] / 2.6)}
Fw = (230x) + (86N) ---Eq1

ΣF = 0
Vertical Forces
Fn + Wm + WL = 0
Fn + [-(600N)] + [-(300N)] = 0
Fn = (600N) + (300N)
Fn = 900N
Horizontal Forces
Ff + Fw = 0 --- Eq2

Ff = µFn
Ff = 0.4 X 900N
Ff = 360N

From Eq2
(360N) + (-Fw) = 0
Fw = 360N

From Eq1
Fw ={[(600N X xm) + (300N X1.5m Sin 30°m)] / (3m Cos 30°m)}
360N ={[(600N X xm) + (300N X1.5m Sin 30°m)] / (3m Cos 30°m)}
(360N) (3m Cos 30°m) = {[(600N X xm) + (300N X1.5m Sin 30°m)]}
{[(360N) (3m Cos 30°m) - (300N X1.5m Sin 30°m)] / (600N)} = xm
x = 1.18m

Length of AP = x / Cos 60°
Length of AP = 2.36m
He can climb 80% of the ladder.

     
Worksheet 1 Static Equilibrium
Worksheet 1 Static Equilibrium Answers
Worksheet 2 Mechanical Properties of Solids
Worksheet 2 Mechanical Properties of Solids Answers
 
ladder

6. The system in Fig. 6.1 is in equilibrium with the string in the center exactly horizontal.
Find (a) tension TL, (b) tension Th, (c) tension Tr and (d) angle θ.

Fig 6.1.
Answer  
ladder

Since X and Y are in equilibrium:
∑F = 0
At X
∑Horizontal forces = 0
[-(TL Sin 35°)] + Th = 0 ---Eq1
∑Vertical forces = 0
(-80.0N) + (TL Cos 35°) = 0 ---Eq2

At Y
∑Horizontal forces = 0
Tr Sin θ + (-Th) = 0 ---Eq3
∑Vertical forces = 0
(-100N) + Tr Cos θ° = 0 ---Eq4

From Eq1
Th = TL Sin 35° ---Eq5


From Eq2
TL = (80.0N)/(Cos 35)
TL = 97.7N ---Eq6

Eq5 becomes Th = (97.7N) Sin 35°
Th = 56.0N

From Eq3
Tr Sin θ = Th
Tr Sin θ = 56.0N ---Eq7

From Eq4
Tr Cos θ = 100N ---Eq8

Dividing Eq7 by Eq8
Tan θ = (56.0N)/(100N)
θ = 29.3 °

From Eq8
Tr= (100.0N)/(Cos 29.3 °)

Tr= 114N

     
Worksheet 1 Static Equilibrium
Worksheet 1 Static Equilibrium Answers
Worksheet 2 Mechanical Properties of Solids
Worksheet 2 Mechanical Properties of Solids Answers
 
ladder

7. The crane in Fig. 7.1 is in equilibrium. A ball of mass 225 kg hangs from the end of the uniform strut whose mass is 50.0 kg.
Find (a) the tension Tc in the cable and the (b) horizontal and (c) force exerted on the strut by the hinge.

Fig 7.1.
Answer

Weight of Strut = 500N
Weight of ball = 2250kg
Length of Strut = L m

The Strut is in Equilibrium so
Στ = 0
Taking moments about F:
(R X 0) + (Tc X BC) + [-(Wb X DF) + [-(Ws X EF) = 0
(0) + (Tc X L Sin 15°) + [-(2250 X L Cos 60°)] + [-(500 X ½L Cos 60°)] = 0
(Tc X L Sin 15°) = (2250 X L Cos 60°) + (500 X ½L Cos 60°)
Tc = [(2250 X Cos 60°) + (500 X ½ Cos 60°)] / (Sin 15°)
Tc = 4830N

The Strut is in Equilibrium so
ΣF = 0
Vertical forces
(Fv) + [-(Wb) + [-(Ws)] + [-(Tc Sin 45°)] = 0
(Fv) = [(Wb) + (Ws)] + (TcSin 45°))
(Fv) = [(Wb) + (Ws)] + (TcSin 45°))
(Fv) = [(2250N) + (500N) + (3415N)]
(Fv) = 6165N
Horizontal forces
(Fh ) + [-(Tc Sin 45 °)]] = 0
(Fh ) = (Tc Sin 45 °)
(Fh) = 3415N
F2 = Fv2 + Fh2
F = 7048N
Tan Θ = Fv / Fh
Θ = 61 °

F = 7048N at elevation 61 °.

ladder
     
Worksheet 1 Static Equilibrium
Worksheet 1 Static Equilibrium Answers
Worksheet 2 Mechanical Properties of Solids
Worksheet 2 Mechanical Properties of Solids Answers
 
ladder

8. The stepladder shown in Fig. 8.1, has sides AC and CE are each 3.0 m long with a hinge at C. The tie bar BD is 0.6m long, halfway up. An man of mass 80kg climbs 2.0m along the ladder. Assuming that the floor is frictionless and neglecting the weight of the ladder, find (a) the tension in the tie–rod and the forces exerted on the ladder by the floor at (b) A and (c) E.

Fig 8.1.
Answer  
 
 
     
Worksheet 1 Static Equilibrium
Worksheet 1 Static Equilibrium Answers
Worksheet 2 Mechanical Properties of Solids
Worksheet 2 Mechanical Properties of Solids Answers
 

9. A 50 kg wooden crate rests on a wooden ramp with an adjustable angle of inclination and a coefficient of friction between the block and ramp of 0.30.
(a) Draw a free body diagram of the crate.
(b) If the angle of the ramp is set to 10°, determine…

(i) the component of the crate's weight that is perpendicular to the ramp
(ii) the component of the crate's weight that is parallel to the ramp
(iii) the normal force between the crate and the ramp
(iv) the static friction force between the crate and the ramp

(c) At what angle will the crate just begin to slip?

Answer (a)  
ladder

W = Weight of crate
W = mg
W = 500N
FN = Force normal to the plane
F = Frictional force (oposing the tendency of the block to move down the plane)

(i) W = W Cos 10°
W = 492N

(ii) W= W Sin 10°
W= 86.8N

(iii) System is in equilibrium so ΣF = 0
-(W) +  FN = 0
FN = 482N

(iv) System is in equilibrium so ΣF = 0
-(W) +  F = 0
F = 86.8N

(iv) The block will begin to slip when F = Fmax
FN = W
So FN = W Cos θ---Eq1
And F = W
So F = W Sin θ ---Eq2
But F = µWand W = FN
Fmax = µFN
Dividing Eq2 by Eq1
(Fmax) / (FN) = Tan θ
Since Fmax =
Tan θ = (µFN) / (FN)
Tan θ = µ
θ = 16°
If θ = 16° then W Sin θ > µW Cos θ
When tan θ > µ the block will slip down the plane.

     
Worksheet 1 Static Equilibrium
Worksheet 1 Static Equilibrium Answers
Worksheet 2 Mechanical Properties of Solids
Worksheet 2 Mechanical Properties of Solids Answers
 
 

10. A farmer has a uniform bar which can swing out on a hinges to secure his access road. The 5m bar has a mass of 20kg and is supported by a cable hinged to the top of the gate post and attached to the middle of the bar. A second hinge is attached to the post at its mid point of the 2m post. Calculate
(a) the tension, T, in the cable.
(b) the force on the hinge at the end of the bar.

Fig 10.1.
Answer

W = Weight of the bar
W = 200N

The bar is in equilibrium so:
ΣFBAR = 0
Tan θ = 1/2.5
θ = 21.8 °

Vertical forces on the bar
ΣFv = 0
[-(W)] + T Cos θ = 0
T = W/Cos θ
T = 215N

Horizontal forces on the bar
ΣFH = 0
R + [-(T Sn θ)] = 0
R = 80N

ladder
 
     
<
>
Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.