A block of mass $$\(m \mathrm{~kg}\)$$ is held in equilibrium below a horizontal ceiling by two strings, as shown in the diagram. One of the strings is inclined at $$\(45^{\circ}\)$$ to the horizontal and the tension in this string is $$\(T \mathrm{~N}\)$$. The other string is inclined at $$\(60^{\circ}\)$$ to the horizontal and the tension in this string is $$\(20 \mathrm{~N}\)$$. Find $$\(T\)$$ and $$\(m\)$$. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................
Exam No：9709_w20_qp_42 Year：2020 Question No：3
Answer:
$20 \cos 60=T \cos 45$
$T=10 \sqrt{ } 2$ or $T=14.1$
$20 \sin 60+T \sin 45=m g$ or $W$
$20 \sin 60+T \sin 45=m g$
$m=2.73[=\sqrt{ } 3+1]$
Alternative method for question 3
$\begin{array}{l}{\left[\frac{T}{\sin 150}=\frac{m g \text { or } W}{\sin 75}=\frac{20}{\sin 135}\right]} \\ \frac{T}{\sin 150}=\frac{m g}{\sin 75}=\frac{20}{\sin 135} \\ \text { Attempt to solve for either } T \text { or } m \text { or } W \\ T=10 \sqrt{ } 2 \text { or } T=14.1 \\ m=2.73[=\sqrt{ } 3+1]\end{array}$
Alternative method for question 3
$
\begin{array}{c}
{\left[\frac{T}{\sin 30}=\frac{m g \text { or } W}{\sin 105}=\frac{20}{\sin 45}\right]} \\
\frac{T}{\sin 30}=\frac{m g}{\sin 105}=\frac{20}{\sin 45}
\end{array}
$
Attempt to solve for either $T$ or $m$ or $W$
$T=10 \sqrt{ } 2$ or $T=14.1$
$
m=2.73[=\sqrt{ } 3+1]
$
$T=10 \sqrt{ } 2$ or $T=14.1$
$20 \sin 60+T \sin 45=m g$ or $W$
$20 \sin 60+T \sin 45=m g$
$m=2.73[=\sqrt{ } 3+1]$
Alternative method for question 3
$\begin{array}{l}{\left[\frac{T}{\sin 150}=\frac{m g \text { or } W}{\sin 75}=\frac{20}{\sin 135}\right]} \\ \frac{T}{\sin 150}=\frac{m g}{\sin 75}=\frac{20}{\sin 135} \\ \text { Attempt to solve for either } T \text { or } m \text { or } W \\ T=10 \sqrt{ } 2 \text { or } T=14.1 \\ m=2.73[=\sqrt{ } 3+1]\end{array}$
Alternative method for question 3
$
\begin{array}{c}
{\left[\frac{T}{\sin 30}=\frac{m g \text { or } W}{\sin 105}=\frac{20}{\sin 45}\right]} \\
\frac{T}{\sin 30}=\frac{m g}{\sin 105}=\frac{20}{\sin 45}
\end{array}
$
Attempt to solve for either $T$ or $m$ or $W$
$T=10 \sqrt{ } 2$ or $T=14.1$
$
m=2.73[=\sqrt{ } 3+1]
$
Knowledge points:
4.1.1 identify the forces acting in a given situation； e.g. by drawing a force diagram.
4.1.2 understand the vector nature of force, and find and use components and resultants； Calculations are always required, not approximate solutions by scale drawing.
4.1.3 use the principle that, when a particle is in equilibrium, the vector sum of the forces acting is zero, or equivalently, that the sum of the components in any direction is zero （Solutions by resolving are usually expected, but equivalent methods (e.g. triangle of forces, Lami’s Theorem, where suitable) are also acceptable; these other methods are not required knowledge, and will not be referred to in questions.）
Solution:
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