# Physics 9702 Doubts | Help Page 173

__Question 863: [Dynamics > Newton’s laws of motion]__
A body experiences a varying
resultant force that causes its momentum to vary, as shown in graph.

At which point does the resultant
force have largest value?

**Reference:**

*Past Exam Paper – November 2014 Paper 13 Q10*

__Solution 863:__**Answer: B.**

From Newton’s 2

^{nd}law, the resultant force is the rate of change of momentum.
Resultant force F = Δp / t

In a momentum-time graph, the
resultant force is given by the gradient of the graph.

The resultant force will have the
largest value at the point where the gradient has the largest value. This is
indicate by the steepness of the tangent at that point. The steeper the
tangent, the greater is the gradient and hence the resultant force.

Of the 4 points, the tangent is
steepest at point B.

__Question 864: [Dynamics > Equilibrium]__**(a)**Define

*centre of gravity*.

**(b)**A uniform rod AB is attached to vertical wall at A. Rod is held horizontally by a string attached at B and to point C, as shown in Fig.1.

Angle between the rod and the string
at B is 50°. The rod has length 1.2 m and weight 8.5 N. Object O of mass M is
hung from the rod at B. Tension T in the string is 30 N.

(i) Use resolution of forces to
calculate the vertical component of T.

(ii) State

*principle of moments*.
(iii) Use principle of moments and
take moments about A to show that weight of object O is 19N.

(iv) Hence determine mass M of the
object O.

**(c)**Use concept of equilibrium to explain why a force must act on the rod at A

**Reference:**

*Past Exam Paper – June 2013 Paper 22 Q3*

__Solution 864:__**(a)**Centre of gravity is defined as the point where (all) the weight (of the body) is considered / seems to act.

**(b)**

(i) Vertical component of T =
30cos(40) = 22.98 ≈ 23N

(ii) The principle of moments states
that the

__sum__of the clockwise moments about a__point__equals the__sum__of the of the anticlockwise moments (about the same point)
(iii)

Let W be the weight of object O

Sum of clockwise moments about A =
Sum of anti-clockwise moments about A

(8.5 ×
0.60) + (1.2 × W) = 23 × 1.2

1.2W = 27.6 – 5.1

W = 18.75 ≈ 19N

{Here is a suggested
solution for Q3(b)(iii):

The previous part (vertical
component of T) is equal to 23 N

Now my method of solving
it is:

Tan 50 = AC / 1.2

AC = 1.43

23 * 1.43 = (8.5+W) * 1.2

W = 18.9 = 19.0 N

Now the method in the
marking scheme is a bit different, but my final answer ( 19.0 ) is still the
same. Is my method acceptable or not?

Why this method is not
appropriate, even if the correct answer is obtained?

1st mistake:

Moment is the product of
force and the perpendicular distance of the force from the pivot. Here BOTH the
distance AC and the force are VERTICAL. This does not have any turning effect
at A.

2nd mistake:

The centre of gravity does
not act at 1.2m from the pivot BUT at the CENTRE.}

(iv)

W = Mg

So, M = W / g = 19 / 9.81 = 1.9 kg

**(c)**For equilibrium, the resultant force and moment must be zero. A force must act on the rod at A since the upward force does not equal downward force

__OR__the horizontal component of T is not balanced by the forces shown.

{The force would start at
point A and its direction would be about north east (the angle with the
horizontal is not necessary 45° - it needs to be determined)}

__Question 865: [Measurement > Uncertainty]__
In an experiment to determine
acceleration of free fall g, the period of oscillation T and length

*l*of a simple pendulum were measured. Uncertainty in the measurement of*l*is estimated to be 4%, and the uncertainty in the measurement of T is estimated to be 1%.
Value of g is determined using the
formula

g = 4π

^{2}*l*/ T^{2}
What is the uncertainty in the
calculated value for g?

A 2% B 3% C
5% D 6%

**Reference:**

*Past Exam Paper – June 2013 Paper 11 Q5*

__Solution 865:__**Answer: D.**

g = 4π

^{2}*l*/ T^{2}
Δg / g = (Δ

*l*/*l*) + 2(ΔT / T)
Percentage uncertainty (Δ

*l*/*l*) = 4%
Percentage uncertainty (ΔT / T) = 1%

Percentage uncertainty in g = 4 +
2(1) = 4 + 2 = 6%

__Question 866: [Matter > Deformation]__
Metal cube of side

*l*is placed in a vice and compressed elastically by two opposing forces F.
How will Δ

*l*, the amount of compression, relate to*l*?
A Δ

*l*α 1 /*l*^{2}B Δ*l*α 1 /*l*C Δ*l*α*l*D Δ*l*α*l*^{2}**Reference:**

*Past Exam Paper – November 2011 Paper 11 Q26 & Paper 13 Q25*

__Solution 866:__**Answer: B.**

Young modulus E = stress / strain =
(Force/Area) / (extension/original length)

Young modulus E = (F/A) / (e/

*l*) = F*l*/ Ae
The Young modulus E is proportional
to

*l*/ Ae.
Area =

*l*^{2}
Compression, e = Δ

*l*
E α

*l*/ (*l*^{ 2}× Δ*l*)
E α 1 / (

*l*× Δ*l*)
So, Δ

*l*α 1 / (E*l*) giving Δ*l*α 1 /*l*since the Young modulus E is a constant.
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