Magnitude and acceleration

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[SOLVED] Magnitude and acceleration

Homework Statement


Three astronauts, propelled by jet backpacks, push and guide a 130 kg asteroid toward a processing dock, exerting the forces shown in the figure, with F1 = 32 N, F2 = 52 N, F3 = 39 N, θ1 = 30°, and θ3 = 60°. What is the (a) magnitude and (b) angle (measured relative to the positive direction of the x axis) of the asteroid's acceleration?


Homework Equations


F=ma


The Attempt at a Solution



If i had to make a guess....would i add the forces up which would be...129=ma where m is 130kg? Or would the magnitude be each mass squared and the square root of those sums which would be 72.45 in this case?
 

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Answers and Replies

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Ok, so i believe i figured out the x component of the net forces which i got to be 99.21 and -y component was -17.77. To get the magnitude for part a, I take 99.21^2 and -17.77^2. Add them together and take the square root which i got to be 100.79. However, i put that answer into the program and its wrong. Any ideas? I also got the x component for the acceleration which is .76 and the y is -.137. Then i take y/x and the inverse tan of that answer?
 
  • #3
PhanthomJay
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Ok, so i believe i figured out the x component of the net forces which i got to be 99.21 and -y component was -17.77. To get the magnitude for part a, I take 99.21^2 and -17.77^2. Add them together and take the square root which i got to be 100.79. However, i put that answer into the program and its wrong. Any ideas? I also got the x component for the acceleration which is .76 and the y is -.137. Then i take y/x and the inverse tan of that answer?
That looks better, but the magnitude of the net force is 100.79...that's not the magnitude of the acceleration. Your acceleration components look good (don't forget the units and round offs), angle theta looks good , ( is it clockwise or counterclockwise from x axis).....so now what do you get for the magnitude and direction of the acceleration?
 
  • #4
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for the magnitude....im guessing youd take each term squared and take teh square root of that sum. Like for before with the magnitude. Im not entirely sure how the 100.79 comes into play then......
 
  • #5
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For part A i got .77 m/s^2 and part b, i got -10.22 degrees but not sure of what to put in. Would i subtract that from 360?
 
  • #6
PhanthomJay
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For part A i got .77 m/s^2 and part b, i got -10.22 degrees but not sure of what to put in. Would i subtract that from 360?
Looks good, the .77m/s^2 acceleration acts about 10 degrees clockwise from the positive x axis, or 350 degrees counterclockwise from the positive x axis. I'd say use .77 at 350, although .77 at -10 would also be correrct. Note that you calculated the net force to be 100.79, so using F_net=ma, you'd get a =100.79/130 = .77, same result, and the angle at which it acts would also be (must be) the same as the acceleation angle (acceleration is always in the direction of the net force).
 

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