# Physics 9702 Doubts | Help Page 228

__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*cos

*θ*– 500 – 200 = 0

*F*cos

*θ*= 700 .............................. (1)

Resolving forces horizontally,

*F*sin

*θ*–

*R*= 0

*F*sin

*θ*=

*R*= 130 ..................... (2)

(1)

^{2}+ (2)^{2}:*F*^{2}cos^{2}*θ*+*F*^{2 }sin^{2}*θ*= 700^{2}+ 130^{2}
{

*F*^{2}(cos^{2}*θ*+ sin^{2}*θ*) but cos^{2}*θ*+ sin^{2}*θ*= 1}*F*= √(700

^{2}+130

^{2}) = 712 N

(2) ÷ (1): tan

*θ*= 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 h

_{1}and h_{2}is possible?
h

_{1}/ cm h_{2}/ 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 = h

_{1}ρg (1)
Similarly, for the right section of
the diagram,

8000 – p = h

_{2}ρg (2)
To eliminate p, take (equation 1 –
equation 2)

(h

_{1}-h_{2})ρg = 8000 Pa
h

_{1 }– h_{2}= 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}
4/O/N/02 Q.5(b)

ReplyDeleteSee solution 1083 at

Deletehttp://physics-ref.blogspot.com/2015/12/physics-9702-doubts-help-page-229.html

great work!

ReplyDeleteWhy is the pressure at P subtracted from the pressure at bulbs in Q1080? Thanks :)

ReplyDeletethe unequal length of column of liquid in the pipe indicates that there is a difference in pressure between P and the bulbs and this difference in pressure is given by P = h rho g

Delete