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Question 1079: [Forces > Static Equilibrium]

A ladder of length 10 m and weight 200 N leans against a smooth wall such that it is at an angle of 60° to the horizontal. A boy of weight 500 N stands on the ladder ¼ of the way from its lower end. Calculate the normal reaction at the wall and the magnitude and direction of the resultant force acting on the lower end of the ladder.

Reference: ‘Pacific Physics A Level Volume 1,’ by POH LIONG YONG, PG 48 Example 4.2

Solution 1079:

The ladder is in equilibrium. This means that the resultant force and the resultant moment on the ladder are zero.

Since the wall is smooth {there is no friction to oppose the motion of end B of the ladder in contact with the wall}, the reaction R is normal to the wall.

The forces acting on the lower end A of the ladder consists of the normal reaction vertically upwards and the friction in the direction AO {as end B tends to move down against the wall, end A of the ladder would tend to move to the left. Thus friction of the ladder with the ground, which opposes motion, would be to the right}. Therefore, the resultant force F at A is in the direction as shown.

Since the ladder is in equilibrium, the algebraic sum of the moments of forces about any axis is zero {the resultant moment is zero}.

Taking moments about A (so that the force F does not appear in the equation {as its distance from the pivot is zero – it is actually at the pivot itself}),

{The perpendicular distance {which is vertical} of R from the pivot A is OB. Length of the ladder, AB = 10m. Thus, sin60° = OB / 10. This causes an anticlockwise moment.}

{The weight (200N) of the ladder acts at the middle (at a distance of 5m along the ladder from A). Considering the triangle formed up to this point, the perpendicular distance {which is horizontal} of the weight is 5 cos60°.}

{The boy, of weight 500N, is at ¼ of the way from end A of the ladder. This is equal to a distance of ¼ (10) = 2.5m from end A along the ladder. Again, considering the triangle formed up to this point, the perpendicular distance {which is horizontal} of the weight is 2.5 cos60°.}

{The 2 weights (200N and 500N) cause a clockwise moment.}

R (10 sin60°) − 200 (5 cos60°) − 500 (2.5 cos60°) = 0

10 R sin60° = 2 250 cos60°

Normal reaction, R = 130 N

Also, the algebraic sum of the components of forces along any direction is zero {the resultant force is zero for equilibrium}. Therefore, if θ is equal to the angle F makes with the vertical, resolving forces vertically,

F cosθ – 500 – 200 = 0

F cosθ = 700 .............................. (1)

Resolving forces horizontally,

F sinθ – R = 0

F sinθ = R = 130 ..................... (2)

(1)² + (2)² : F² cos²θ + F² sin²θ = 700² + 130²

{F²(cos²θ + sin²θ)       but       cos²θ + sin²θ = 1}

F = √(700²+130²) = 712 N

(2) ÷ (1): tanθ = 130 / 700

θ = 10° 31′ from the vertical

Question 1080: [Pressure > Liquid]

Two bulbs X and Y containing air at different pressures are connected by a tube P which contains two mercury threads.

The density of mercury is 13 600 kg m^–3.

Which pair of values of h₁ and h₂ is possible?

h₁ / cm             h₂ / cm

A         4.0                   2.0

B         6.0                   6.0

C         12.0                 18.0

D         18.0                 12.0

Reference: Past Exam Paper – June 2013 Paper 12 Q20

Solution 1080:

Answer: D.

The answer to this question cannot reliably be obtained by guesswork.

Pressure difference = hρg       where h is the difference in levels of the liquid

Let pressure at P = p.

Considering the left section of the diagram,

16000 – p = h₁ρg                     (1)

Similarly, for the right section of the diagram,

8000 – p = h­₂ρg                       (2)

To eliminate p, take (equation 1 – equation 2)

(h₁-h­₂)ρg = 8000 Pa

h₁ – h­₂ = 0.060m

Choice D is the only set that fits this equation.

Question 1081: [Matter > Deformation of Solids > Hooke’s law]

A spring is compressed by a force. The graph shows the compressing force F plotted against the length L of the spring.

What is the spring constant of this spring?

A 0.2 N m^–1                 B 5 N m^–1                    C 100 N m^–1                D 200 N m^–1

Reference: Past Exam Paper – June 2010 Paper 11 Q20 & Paper 12 Q19 & Paper 13 Q21

Solution 1081:

Answer: D.

Hooke’s law: F = - kx

Spring constant, k = - F/x = - gradient

Consider points (50,10) and (90,2).

Gradient = (10 – 2) / [(50 – 90)×10^-3] = - 200

Spring constant, k = 200 Nm^-1