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 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
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
DeleteI have benefited from this page, thank you.
ReplyDelete