Question

In the following figure, an object of mass 1.2Kg is at rest at point P. If R and F are the reaction and the frictional force, respectively, then

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Answer 1

Explanation:

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Answer 2

Given:

mass  = 1.2Kg

g = 10

R and F are the reaction and the frictional force

$\begin{array}{l}R=m g \cos 30^{\circ} \\\\=1.2 \times 10 \times \frac{\sqrt{3}}{2}\end{array}$

$R=(6 \sqrt{3}) N$

$\begin{array}{l}f=m g \sin 30^{\circ} \\\\=1.2 \times 10 \times \frac{1}{2}\end{array}$

=1.2 × 5

F = 6

Hence the value of R is $R=(6 \sqrt{3}) N$  and value of F is F = 6