confusing question?

[TMJ]

A 100kg trolley is being pushed up a rough 30degree incline by a constant force F. The frictional force Ff between the incline and the trolley is 110N.
Determine the value that will move the trolley up the incline at a constant velocity of 5ms^-1. 
Answer = 600 N
 
When I calculated it I did    F=100 x 9.8sin(30)-110  as friction opposes direction of motion. this gave me an answer of 380N
However adding the friction gives you the answer of 600N. Why is this? It confuses me as I though you must consider the direction of all forces.

[Hancock]

If we want a constant velocity, the net force must equal 0. Netwon's Second Law states that 0 net force = 0 acceleration. A constant velocity has no acceleration by definition.

Therefore, if we consider the forces going up the plane as positive, we have

Fnet = F(push) - F(friction) - F(weight component)
Fnet = F - 100 - mg*sin(30)
0 = F - 100 - 100*10*0.5
F = 100 + 100*10*0.5 = 600 N

[FlorianK]

A 100kg trolley is being pushed up a rough 30degree incline by a constant force F. The frictional force Ff between the incline and the trolley is 110N.
Determine the value that will move the trolley up the incline at a constant velocity of 5ms^-1. 
Answer = 600 N
 
When I calculated it I did    F=100 x 9.8sin(30)-110  as friction opposes direction of motion. this gave me an answer of 380N
However adding the friction gives you the answer of 600N. Why is this? It confuses me as I though you must consider the direction of all forces.

Ok, constant velocity means the Net-Force is zero.
--> 0=F-110-9.8*sin(30)*100
110+9.8*sin(30)*100=F
F=110+490
F=600

[TMJ]

Why did you make the weight force a negative?

[Phy124]

They took up the slope as the positive direction. Therefore, the component of the weight force along the slope will be negative, as it acts down the slope.

[TMJ]

This question involves a good understanding of the concepts more than the maths involved in my opinion.

[Hancock]

This question involves a good understanding of the concepts more than the maths involved in my opinion.

I'll chuck a picture up in a sec.

[TMJ]

thanks

[Hancock]

As we can see, the component of the weight force points down the plane and has a magnitude . Since we defined the up the plane as positive, we have the equation:



And Florian's and my working out runs from there.