Q.321. Mechanics - Static equilibrium involving friction.

Question 321.

Refer to the figure as above.
A uniform ladder with mass m₂ and length L rests against a smooth wall. A do-it-yourself enthusiast of mass m₁ stands on the ladder at a  distance d from the bottom (measured along the ladder). The ladder makes an angle θ with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude ƒ between the floor and the ladder. N₁ is the magnitude of the normal force exerted by the wall on the ladder, and N₂ is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive.

1)
What is the minimum coeffecient of static friction μ_min required between the ladder and the ground so that the ladder does not slip? Express μ_min in terms of m₂, m₁, d , L , and θ.
2)
Suppose that the actual coefficent of friction is one and a half times as large as the value of μ_min. That is, μ_s= (3/2) μ_min. Under these circumstances, what is the magnitude of the force of friction ƒ that the floor applies to the ladder? Express your answer in terms of m₁, m₂, d , L ,g and θ.

Answer 321.
1)
Balancing vertical forces,
N₂ = m₁ + m₂
 When the ladder is on the verge of slipping, frictional force from right to left at the bottom of the ladder with the floor,
F = μ_min * N₂ = μ_min * (m₁ + m₂)
 Taking moments about the point of contact of the ladder with the wall,
N₂ * Lcosθ - F * Lsinθ - m₁ * (L - d) cosθ - m₂ * (L/2) cosθ = 0
=> (m₁+m₂) * Lcosθ - μ_min * (m₁+m₂) * Lsinθ - m₁ * (L - d) cosθ - m₂ * (L/2) cosθ = 0
=> μ_min * (m₁+m₂) * Lsinθ = (m₁+m₂) * Lcosθ - m₁ * (L - d) cosθ - m₂ * (L/2) cosθ
=> μ_min * (m₁+m₂) * Lsinθ = m₁ * dcosθ + (m₂/2) Lcosθ

=> μ_min
= (dm₁ + Lm₂/2)cotθ / [L(m₁+m₂)]
= [(d/L)m₁ + (1/2)m₂] cotθ / (m₁+m₂).

2)
Balancing vertical forces,
N₂ = m₁ + m₂
Balancing horizontal forces,
f = N₁
Taking moments of all the forces about the point of contact of the ladder with the floor,
- N₁ Lsinθ + m₂g (L/2)cosθ + m₁g dcosθ = 0
=> f = N₁ = (m₂/2 + m₁d/L) gcotθ.

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Q.320. Geometry

Question 320.
Find AG in the figure as under.

Answer 320.
As the answer is independent of the radius of the circle, the best choice of the radius will be such that DC passes through the center of the circle. As one vertical line is at a distance 4 from A and the other at a distance 9 from A, the center must be at a distance (1/2) * (4 + 9) = 6.5 from A
=> radius of the circle can be taken as 6.5.
 Refer to the figure drawn.

Δs AOC and BOD are isosceles and congruent.
[OA = OB = radius and OC = OD = radius and vertically opposite angles O are equal.]
=> x
= DF
= √[(6.5)^2 - (2.5)^2]
= 6.

Though  my  above  answer  was  selected   as  the  best, it  involves  an  unproved  assumption  of the  answer being  independent  of  the  radius  of  the  circle.

One  of  my  contacts,  Rakesh  Dubey,  who  is  very  good  at  solving  challenging  problems of  geometry  besides those  of  other  topics  of  maths,  provided  a  superior  solution  in  which  he  has  solved  the  problem  without  any  assumption  as  above.  He  had  posted  his  solution  in  comments  section  after the question  was  closed  for  answering.  His  solution  is  as  under.

Refer  to  the  following  figure.

 

Join  A  and  B  to  C  and  D.
In  right  triangle  ADB,  it  can  be  proved  with  simple  geometry  that
DF^2  =  AF * FB   ...   ...   ...   ...   ...   ( 1 )

In  right  Δs  CAG  and  BDF,
angle  ACG  =  angle  DBF
=> Δs  CAG  and  BDF  are  similar
=> AC / BD = AG / DF = x / DF
=> x = (AC * DF) / BD
=> x^2 = (AC^2 * DF^2) / BD^2   ...   ( 2 )

In  right  Δs  CAE  and  BAC,
angle A is common
=> Δs  CAE  and  BAC are similar
=> AC/BA = AE/AC
=> AC^2 = AE * BA   ...   ...   ...   ...   ( 3 )

In  right  Δs  BDF  and  BAD,
angle B is common
=> Δs  BDF  and  BAD are similar
=> BD^2 = AB * BF   ...   ...   ...   ...  ( 4 )

Plugging DF^2, AC^2 and BD^2 from ( 1 ), ( 3) and ( 4 ) in eqn. ( 2 ),
x^2 = [AE * BA * AF * FB) / (AB * BF)
=> x^2 = AE * AF = 4 * 9 = 36
=> x = AG = 6.

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Q.319. Geometry

Question 319.
The figure shows a triangle ABC. The incircle O is tangent to AC at D. M is the midpoint of AC. Line AE is perpendicular to BO extended. Prove that the measure of DM is equal to half the measure of CE.

Answer 319.

In right Δs ABL and EBL,

side BL is common and

∠ABL = ∠EBL
=> Δs ABL and EBL are congruent
=> BE = AB = c
=> CE = BC - BE = a - c ... (1)
 DM
= AM - AD
= b/2 - rcot(A/2)
= b/2 - (s - a)
= b/2 - [(a+b+c)/2 - a]
= - a/2 - c/2 + a
= (a - c)/2 ...   ...   ...   ...    (2)

From (1) and (2),
DM = (1/2) CE.

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Q.318. Mechanics - Friction

Question 318.
A 29.5 kg block is connected to an empty 1.00 kg bucket by a cord running over a frictionless pulley. The coefficient of static friction between the table and the block is 0.485 and the coefficient of kinetic friction between the table and the block is 0.320. Sand is gradually added to the bucket until the system just begins to move.
(a) Calculate the mass of sand added to the bucket.
(b) Calculate the acceleration of the system. (downward).

Answer 318.
Maximum static frictional force on the block
= μ(s) * mg
= (0.485) * (29.5) * (9.81) N
≈ 140.4 N.

(a)
Let m = mass of sand to be added in kg to start the motion of the block
=> (m + 1) * 9.81 = 140.4
=> m ≈ 13.31 kg

(b)
If T = tension in the cord once the motion starts, then
acceleration of the bucket = acceleration of the block = a
=> (13.31 + 1) * 9.81 - T = (13.31 + 1) a ... (1)
and T - (29.5) * (9.81) * (0.320) = 29.5 a ... (2)
Adding (1)and (2),
(13.31 + 1) * 9.81 - (29.5) * (9.81) * (0.320) = (13.31 + 1) a + 29.5a
=> 43.81 a = 47.77
=> a = 1.09 m/s^2.

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Q.317. Geometry

Question 317.
Refer to the following geometric figure and prove that
(A1 + A2) - (A3 + A4) = 8A5.

Answer 317.

Refer to the following figure.

Let P be a point on GC such that GP = GE.

Draw a horizontal line through P forming a chord of the circle.
FQ is a vertical line through F meeting this chord and SR is another vertical line such that FS = EF
Thus, PRSE is a rectangle containg four equal rectangles each of area 2 * A5
=> area PRSE = 8 * A5 and is a part of A1.

Also, area above the horizontal chord throgh P and on right of CP = A4 and is also part of A1
CP = ED => area above the horizontal chord through P and on left of CP = A2
AE = SB => area between the horizontal chords on left of PE = raea on right of RS = A3 - A2 which is also part of A1.

=> A1 is made up of 8 * A5 + A4 + (A3 - A2)
=> (A1 + A2) - (A3 + A4) = 8 * A5.

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Q.316.Probability - Application of Bayes' Theorem

Question 316.
Three urns contain colored balls at follows:
Urn One: 3 red, 4 white, 1 blue
Urn Two: 1 red, 2 white, 3 blue
Urn Three: 4 red, 3 white, 2 blue
One urn is chosen at random and a ball is withdrawn. It turns out to be red. What is the probability that is came from Urn Two?

Answer 316.
Let Ui = the event that urn i is selected
and R = the event that the red ball is selected from the selected urn

As the urns are selected randomly, P(U1) = P(U2) = P(U3) = 1/3

=> P(R/U1) = No. of red balls in urn 1 / total no. of balls in urn 1
                           = 3/8
Similarly, P(R/U2) = 1/6 and P(R/U3) = 4/9

By Bayes' Rule,
P(U2/R)
= P(U2) * P(R/U2) / [P(U1) * P(R/U1) + P(U2) * P(R/U2) + P(U3) * P(R/U3)]
= (1/3) * (1/6) / [(1/3) * (3/8) + (1/3) * (1/6) + (1/3) * (4/9)]
= (1/6) / (3/8 + 1/6 + 4/9)
= (1/6) * (72/71)
= 12/71.

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Q.315. Geometry

Question 315.
Find the angle α in the following figure.

Answer 315.

Refer to the figure (link below)

OD is parallel to BC

=> ∠ DOF = ∠ BCO = C/2 [corresponding angles] ... (1)

AD and AE are tangents to the incircle from A

=> AD = AE

=> Δ ADE is isosceles

and ∠ ADE = ∠ AED = (1/2) (π - A) = π/2 - A/2

=> ∠ ODF = π/2 - (π/2 - A/2) = A/2 ... (2)

α is an external angle of Δ DOF

=> α

= sum of the opposite internal angles

= ∠ DOF + ∠ ODF

= C/2 + A/2 ... [From (1) and (2)]

= π/2 - B/2

= 45°.

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Q.314. Minimising cost

Question 314.
A rectangular garden, area of 300 m^2. The cost of fencing three sides is $9 /m and the cost of fencing the fourth side is $15/m. Find the dimensions of the garden such that the cost of fencing is minimum and the minimum cost of fencing the garden.

Answer 314.
Let the two adjacent sides of the fence be of lengths = a and b in m.
=> ab = 300 ... ( 1 ) and
Cost of fencing,
C = 9 * (2a + b) + 15b = 18a + 24b
Plugging a = 300/b in the equation of cost,
C = 5400/b + 24b
For C to be minimum, dC/db = 0 and d^2C/db^2 > 0
dC/db = 0 => - 5400/b^2 + 24 = 0 => b = 15 m and a = 20 m
d^2C/db^2 = 10800/b^3 > 0
=> Minimum cost of fencing will be when one side is 15 m and the other 20 m with cost of fencing one 15 m side being $ 15/m and for the remaining three sides $ 9/m
and the minimum cost
= $ (18* 20 + 24 * 15)
= $ 720.

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