Verified answer

The particle Q is taken into account because the

weight of Q tends to pull the right side of the

string down, with force 0.3*g Newtons, but the

weight of P tends to pull the left side of the

string down, with force 0.4*g Newtons.

Since they are connected by the string, these

pull against eachother. If the plane was removed

and P was resting on your hand, you would only

feel the weight of 0.1 kg pressing down.

It is as if you replaced the two weights with a

single weight of 0.1 kg at P. If freely suspended,

the force 0.1g Newtons would act downwards,

but by being in contact with the plane, the effects

of the force become a tendency to slide parallel

to the plane and also a tendency to push

perpendicularly into the plane.

Any force F acting at say x degrees to the horizontal

has a horizontal component of F cos x,

and a vertical component of F sin x

If you do not understand that bit, look at this webpage

http://www.hk-phy.org/contextual/mechanics/for/ad_...

Extend the string line with a short dotted line and make a dot.

Draw another line from the dot to the slope making a right angle to the slope.

The smaller angle is 30 degrees

The component parallel to the slope is 0.1g cos 30 = 0.1gsin 60

The component parallel to the slope is 0.1g sin 30 = 0.1gcos 60

I hope it makes sense to you now. If not read through again slowly.

Regards - Ian