Physics Dynamics Question homework help

huzjm

1. The problem statement, all variables and given/known data
Find the acceleration in the system of a fletcher’s trolley given m1 = 1.0 kg and m2 = 9.0 kg and a 50 N force of friction exists.
θ = 33°

See the image below, to explain the diagram. This one has an angle of 33 degrees that is why it is hard.

http://i39.tinypic.com/3497zo9.png


2. Relevant equations
I am not sure.

3. The attempt at a solution
Okay so first I found out the horizontal component of m1 by the formula cos(33)*mass(which is 1.0kg) and got the value 0.838 , then I put the formula Fnet=[(M2*A)-Ff]/(m1+m2)
which means 9.0kg*9.81m/s2 - 50 N divided by (0.838+9) .. the answer was 3.9 M/s2 but this is the wrong answer, i asked my teacher, he said you are somewhere close but this is not right. I cannot figure this out, I have been trying different things since hours.

tiny-tim

hi huzjm! welcome to pf!

(try using the X₂ button just above the Reply box )
sorry, but even if you mean a = Fnet/(m1+m2) = [(M2*A)-Ff]/(m1+m2), that's still completely wrong

try calling the tension "T", and doing two F = ma equations (one for each block)

alternatively, if you're treating the two blocks as a single system, your m in ma has to be the total (unadjusted) mass, and you have to use the component of m₁g parallel to the string

try it both ways … what do you get?

huzjm

Thank you very much :) In fact, thank you really very much :)
This is what I did
Fnet = ma = 9kg * a = mg - T = 9kg * 9.8m/s² - T = 88.2N - T
→ T = 88.2 - 9a

For the lighter mass,
Fnet = ma = 1kg * a = T - mgsinΘ - Ff = T - 1kg * 9.8m/s² * sin33º - 50N
1kg * a = T - 5.34N - 50N = T - 55.34N
→ T = a + 55.34

Since T = T,
88.2 - 9a = a + 55.34
32.86 = 10a
a ≈ 3.3 m/s²

Is the answer right now?

tiny-tim

excellent!

(btw, you'll notice you could get the same result by treating it as a one-dimensional motion, with a single body with a mass of 10 kg, friction of 50 N and gravitational forces of 9g N and -gsin33° N )

huzjm

Oh so the formula should have been acceleration= [(9.0kg*9.81m/s) - 50N - (9.81m/s * sin 33°)] / 10 KG

Thank you really very much :)

huzjm

I mean m/s(square)