A block of mass $$\(5 \mathrm{~kg}\)$$ is placed on a plane inclined at $$\(30^{\circ}\)$$ to the horizontal. The coefficient of friction between the block and the plane is $$\(\mu\)$$. When a force of magnitude $$\(40 \mathrm{~N}\)$$ is applied horizontally, in a vertical plane containing a line of greatest slope, the block does not move (see Fig. 6.2). Show that, correct to 3 decimal places, the least possible value of $$\(\mu\)$$ is $$\(0.152\)$$. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_w20_qp_42 Year：2020 Question No：6(b)
Answer:
Attempt to resolve forces parallel to or perpendicular to the incline plane, 3 relevant terms in either direction
$
R=5 g \cos 30+40 \sin 30[=20+25 \sqrt{ } 3=63.3]
$
\(F=40 \cos 30-5 g \sin 30[=20 \sqrt{3}-25=9.64]\)
$
\mu \geqslant 0.152
$
Alternative method for question 6(b)
Attempt to resolve forces horizontally or vertically with 3 relevant terms in either direction
$
40=R \sin 30+F \cos 30[40=1 / 2 R+\sqrt{3} / 2 F]
$
\(5 g=R \cos 30-F \sin 30[5 g=\sqrt{3} / 2 R-1 / 2 F]\)
$
\mu \geqslant 0.152
$
$
R=5 g \cos 30+40 \sin 30[=20+25 \sqrt{ } 3=63.3]
$
\(F=40 \cos 30-5 g \sin 30[=20 \sqrt{3}-25=9.64]\)
$
\mu \geqslant 0.152
$
Alternative method for question 6(b)
Attempt to resolve forces horizontally or vertically with 3 relevant terms in either direction
$
40=R \sin 30+F \cos 30[40=1 / 2 R+\sqrt{3} / 2 F]
$
\(5 g=R \cos 30-F \sin 30[5 g=\sqrt{3} / 2 R-1 / 2 F]\)
$
\mu \geqslant 0.152
$
Knowledge points:
4.1.1 identify the forces acting in a given situation； e.g. by drawing a force diagram.
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
Solution:
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