net force help?

[AllahuAkbar]

A monkey of mass 15kg has escaped backstage. a nearby rope is threaded through an ideal(frictionless massless) Pulley.
Attached to one end of the rope is a 10kg bag of sand. The monkey climbs a ladder and jumps onto the free end of the rope.
The system of the rope and the monkey are now subjected to a net force of F=15 x 9.8 - 10 x 9.8 = 49n. 

Why do you subtract the 10 x 9.8 instead of adding. I thought net force was the sum of all vector forces. 

My suggestion was because the 10 kg sand bag goes in the opposite direction of the monkey it has a negative direction, hence -10 x 9.8 and the monkey has the positive direction as its falling to the ground.

Can someone please correct me.

[LazyZombie]

My suggestion was because the 10 kg sand bag goes in the opposite direction of the monkey it has a negative direction, hence -10 x 9.8 and the monkey has the positive direction as its falling to the ground.

Can someone please correct me.

From the year 11 heinemann book amirite?

Pretty much, you can think of it as the sandbag opposing the greater weight force of the monkey, thus subtracting it, or you can think of each force as a vector and add them together. One force is negative and one is positive as they are going in different directions, and add them.

So, your 'suggestion' is right. But I don't like the way you talk about "positive" and "negative" directions.

You could also do this if we decided upwards was "positive" too. So the monkeys force would be negative, as it would be accelerating downwards, and the sandbags force would be positive as it would be accelerating upwards. Since we decided upwards is positive.

Add the vectors together:
-15x9.8+10x9.8=-49N

 So the total net force would be -49N upwards.
And a negative force just means that the force is in the opposite direction of "upwards".
So -49N upwards would be the same thing as 49N downwards.

Keep in mind that g=10 in the vcaa exam, saves heaps of time.

>< Hope that didn't confuse you.