1. ## Statics

Two forces of 10 kg wt and a third force of P kg wt act on a body in equilibrium. The angle between the lines of action of the 10 kg et forces is 60 degrees.

A particle of mass 5 kg hangs from a fixed point O by a light inextensible string. It is pulled aside by a force P kg wt that makes an angle of 120 degrees with the downward vertical and rests in equilibrium with the string inclined at 60 degrees to the vertical. Find P.

The angles between the forces of magnitude 10 kg wt, 5 kg wt and 5(3)^(.5) {5 'square root' 3) kg wt acting on a particle are 120 degrees and 90 degrees respectively. Show that the particle is in equilibrium.

2. I'm assuming these are three separate questions?

1) Seems incomplete.
2) As the particle is in equilibrium, the resultant of all the forces acting on it will be zero. We can see that the forces acting on the body are its weight, the tension in the string and the force P.

So, resolve the forces in the horizontal and vertical directions.

In the horizontal direction,
$Pcos(30) - Tsin(60) = 0$

Now in the vertical direction, The resultant of $Tcos60$, $Psin30$ and $Weight W$ must be zero.

Now [tex]W=mg[tex] by definition where m is the body's mass (=5kg) and $g=10m/s^2$

So $W=50N$

Now, $Tcos60-50-Psin30=$

but P=T from the first equation.

hence, $Pcos60-50 + Psin30=0$

Hence $2Psin30-50=0$ (since sin30=cos60)

So $P-50=0$

$P=50N$

3)Use the same method. Components of forces in horizontal and vertical direction must add to give zero.