A block of mass $$$5 \mathrm{~kg}$$$ is placed on a plane inclined at $$$30^{\circ}$$$ to the horizontal. The coefficient of friction between the block and the plane is $$$\mu$$$. When a force of magnitude $$$40 \mathrm{~N}$$$ is applied to the block, acting up the plane parallel to a line of greatest slope, the block begins to slide up the plane (see Fig. 6.1). Show that $$$\mu<\frac{1}{5} \sqrt{3}$$$. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mathematics

IGCSE&ALevel

CAIE

Exam No：9709_w20_qp_42 Year：2020 Question No：6(a)

Answer:

$
\begin{array}{l}
R=5 g \cos 30 \quad[=25 \sqrt{ } 3] \\
40-5 g \sin 30-F>0 \\
F=\mu \times 5 g \cos 30 \\
\mu<\frac{1}{5} \sqrt{3}
\end{array}
$
Alternative scheme for question 6(a)
$\begin{array}{l}R=5 g \cos 30[=25 \sqrt{ } 3] \\ 40-5 g \sin 30-F=5 a \\ F=\mu \times 5 g \cos 30 \\ {[40-5 g \sin 30-\mu \times 5 g \cos 30=5 a]} \\ \mu<\frac{1}{5} \sqrt{3}\end{array}$

Knowledge points:

4.1.1 identify the forces acting in a given situation； e.g. by drawing a force diagram.

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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