Fup=Fdown

[physics]

The cart and person=100kg
What is the net frictional force on the cart?
i thought that since Fdown=Fup so i just found F down which is mgsin(angle) =490N

Which arrow represents the net frictional force? (red arrow?)

Which one represents the direction of the force that the slope exerts on the sled? (yellow?)

The rider then releases the brake and the cart accelerates. What is the magnitude of the inital acceleration?
i used the formula a=gsin(angle) = 4.9ms^-2

thanks in advance

[mark_alec]

If Fdown were equal to Fup, then the cart would not slide down at all. Rather, the component of the weight force perpendicular to the slope and the normal force are equal. Once you know the component of the weight force down, you can calculate the friction from the coefficient of friction.

[physics]

If Fdown were equal to Fup, then the cart would not slide down at all. Rather, the component of the weight force perpendicular to the slope and the normal force are equal. Once you know the component of the weight force down, you can calculate the friction from the coefficient of friction.

so its mg? 100 * 9.8 then thats the answer of the fricitonal force?

[Twenty10]

100*9.8 is the weight force (Pink arrow)
There is the normal force (Yellow arrow)
the frictional force is the red arrow.

To calculate friction redraw the triangle. mg is the hyppotonuse. friciton is the opposite of 30 degrees. and the normal is the adjacent of 30 degrees. Therefore friction should be mg*Sin(30).....(from.. Sin(30)=o/h)

then for the release brake part... F=m*a. 490/100=  4.9ms^-2

Fdown does not = Fup as it is on an incline. Fup (The normal force) is = 980*Cos(30).

[physics]

100*9.8 is the weight force (Pink arrow)
There is the normal force (Yellow arrow)
the frictional force is the red arrow.

To calculate friction redraw the triangle. mg is the hyppotonuse. friciton is the opposite of 30 degrees. and the normal is the adjacent of 30 degrees. Therefore friction should be mg*Sin(30).....(from.. Sin(30)=o/h)

but thats Fdown isnt it?

so to find Fup you find F down?

[Twenty10]

sorry i edited my post above...
I dont think that blue arrow actually exists? Like there is no force directly upwards. And uhh/... what does the purple arrow represent? I dont think thats a force either?

[physics]

the net frictional force is the red line? but then why do we have to find Fnormal? is it because F normal is the net fricitonal force?
sorry for my stupididty

[Twenty10]

no we dont need to find normal unless asked for by a question. but u wer saying Fup. So i thought u were refering to the normal...
Nah.In this case the friction opposes the natural motion of the cart. The natural motion being moving down along the inclined plane. (in the direction of the green arrow). So the opposite of this is the red arrow.

All ur answers were right in the first place anyway so there is no stupidity lol. :p

[physics]

Frictional force= to the force going down? to keep it stationary? OMG so i got it..right? thankssss

[Twenty10]

wait. what do you mean by the force going down? do you mean the green arrow or the weight?

coz red arrow = green arrow in terms of magnitude of force.

[/0]

Bit of a dodgy question if you ask me... it assumes that you know how to deal with rolling in vce physics.

Your method of is correct when the cart is stationary. When the wheels start to roll however, you're dealing with something else entirely. First of all, you are given no co-efficient of rolling friction, and second of all, you don't know the size or shape of the wheels. So GG.

[physics]

Bit of a dodgy question if you ask me... it assumes that you know how to deal with rolling in vce physics.

Your method of is correct when the cart is stationary. When the wheels start to roll however, you're dealing with something else entirely. First of all, you are given no co-efficient of rolling friction, and second of all, you don't know the size or shape of the wheels. So GG.

okay i get it thanks btw awesome picture haha

[Stroodle]

Bit of a dodgy question if you ask me... it assumes that you know how to deal with rolling in vce physics.

Your method of is correct when the cart is stationary. When the wheels start to roll however, you're dealing with something else entirely. First of all, you are given no co-efficient of rolling friction, and second of all, you don't know the size or shape of the wheels. So GG.

It is a dodgy question. One second it's a sled, the next second it's a cart...