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

Question

In the following figure, an object of mass 1.2 kg is at rest at point P . lf R and F are the reaction and the frictional force, respectively, then (A)  R = 6 N ; F = 6√3 N  (B)  R = 3 N ; F = 3√3 N  (C)  R = 6 N ; F = 3 N  (D)  R = 6√3 N ; F = 6 N

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e4048fd5e4fd5c8e60a30a3538e04c20-.png" /> Given, m = 1.2 kg R = mg cos 30° = 1.2 × 10 × (√3/2) R = (6√3)N f = mg sin 30° = 1.2 × 10 ×(1/2) F = 6

Q. In the following figure, an object of mass $1.2 k g$ is at rest at point $P$ . lf $R$ and $F$ are the reaction and the frictional force, respectively, then

$R = 6 N$ ; $F = 63 N$

$R = 3 N$ ; $F = 33 N$

$R = 6 N$ ; $F = 3 N$

$R = 63 N$ ; $F = 6 N$

Given, $m = 1.2 k g$
$R = m g cos 3 0^{\circ}$
$= 1.2 \times 10 \times \frac{3}{2}$
$R = (63) N$
$f = m g sin 3 0^{\circ}$
$= 1.2 \times 10 \times \frac{1}{2}$
$F = 6$

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

R=6N ; F=63​N

R=3N ; F=33​N

R=6N ; F=3N

R=63​N ; F=6N