A string is attached to a block of mass $$$4 \mathrm{~kg}$$$ which rests in limiting equilibrium on a rough horizontal table. The string makes an angle of $$$24^{\circ}$$$ above the horizontal and the tension in the string is $$$30 \mathrm{~N}$$$. Find the coefficient of friction between the block and the table. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mathematics

IGCSE&ALevel

CAIE

Exam No：9709_w20_qp_43 Year：2020 Question No：3(b)

Answer:

For resolving horizontally or vertically
$30 \cos 24=F \quad(F=27.406 \ldots)$
$R+30 \cos 24=40 \quad(R=27.797 \ldots)$
$\begin{array}{l}\mu=\frac{30 \cos 24}{40-30 \sin 24} \\ \mu=0.986(0.9859 \ldots)\end{array}$

Knowledge points:

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.）

4.1.4 understand that a contact force between two surfaces can be represented by two components, the normal component and the frictional component

4.1.6 understand the concepts of limiting friction and limiting equilibrium, recall the definition of coefficient of friction, and use the relationship F = nR or F G nR, as appropriate

Answer:

Knowledge points:

Solution:

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