Physics Dynamics Question homework help

AI Thread Summary
The discussion focuses on calculating the acceleration of a system involving two masses, m1 (1.0 kg) and m2 (9.0 kg), with a frictional force of 50 N and an angle of 33 degrees. Initial attempts at finding the acceleration yielded incorrect results, prompting further exploration of the equations of motion and forces acting on the system. Participants suggested using tension in the equations and treating the blocks separately or as a single system for clarity. The correct approach led to an acceleration calculation of approximately 3.3 m/s² after resolving the forces accurately. The final formula for acceleration was confirmed as [(9.0 kg * 9.81 m/s²) - 50 N - (9.81 m/s² * sin 33°)] / 10 kg.
huzjm
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Homework Statement


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


Homework Equations


I am not sure.

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.
 
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welcome to pf!

hi huzjm! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)
huzjm said:
Fnet=[(M2*A)-Ff]/(m1+m2)

sorry, but even if you mean a = Fnet/(m1+m2) = [(M2*A)-Ff]/(m1+m2), that's still completely wrong :redface:

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? :smile:
 
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?
 
excellent! :smile:

(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 :wink:)
 
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 :)
 
I mean m/s(square)
 
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