Question

@ProfBrainstorm Physics

Answer 1

okay, well just supposing the push force was horizontal, we would have basically the same problem as last time, just different numbers

Answer 2

but the question is a little more complicated and we have to take account of that push force being at an angle

Answer 3

In order to work out the net force in the horizontal direction, we need to calculate what is called the horizontal component of the push force
That is given by F * cosine(theta)

Answer 4

Once you have that horizontal component calculated, the rest of the problem goes just the same as before

Answer 5

So 648cos25=587.287446?

Answer 6

yes, 587.3 is probably fine

Answer 7

so the next thing would be to figure out the weight of the mass, then use the drag coefficent to work out the drag force

Answer 8

no, the mass is given as 80kg

Answer 9

the 587.3 is the horizontal component of the push force

Answer 10

ahh makes sense. let me figure it out and tell u what I got

Answer 11

ok

Answer 12

weight is 784.8?

Answer 13

yes

Answer 14

and multiply that by the 0.13 to get the acceleration?

Answer 15

erm, not quite, multiply the weight by 0.13 to get the frictional drag force

Answer 16

okay and subtract that value from the push force?

Answer 17

well you have to be careful here, the drag force acts horizontally, so we need to subtract it from the horizontal component of the push force.

Answer 18

that was the 587.3 that we calculated first

Answer 19

Oh so multiply the weight by 0.13 and subtract that answer from 587.3?

Answer 20

yes, that's right

Answer 21

because forces are vectors, you see, so you always have to think about the components in a particular direction when you add them

Answer 22

okay, that makes sense. I got 485.276 or 485.3

Answer 23

yes, same here, and now you're one step away from getting the acceleration of the mass

Answer 24

push force- drag force? or is it plus?

Answer 25

you've worked out the net force in the horizontal direction, that's what the 485.3 is

Answer 26

Oh i am so sorry!

Answer 27

it's alright

Answer 28

so the 587.3 is the push force and the 102.024 is my drag force. Subtracting those gave me my net force.

Answer 29

yes, in the horizontal direction

Answer 30

that's what I have, so it must be right : )

Answer 31

Thank you very much, I have one more question. I am going to post it on another thread. My teacher isn't very helpful :(

Answer 32

are you happy with this one - the important point is that the force was at an angle, so its effectiveness in the horizontal direction is reduced by the factor cosine(theta)
that was the crucial point

Answer 33

Yeah I understand it, a bit better. It was just an extra step.

Answer 34

ok