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Magnitude of forces

  • Thread starter hoseA
  • Start date
  • #1
61
0
A 2.7 kg mass accelerates at 7 m/s^2 in a
direction 28 degrees north of east. One of the two
forces acting on the mass has a magnitude of
12.6 N and is directed north.
Determine the magnitude of the second
force. Answer in units of N.

I did F2= sqr rt (ma^2+ F1^2)

= 22.715 N

This is apparently incorrect. This seems too easy and yet I cannot figure it out. I'm assuming F2 is in the positive "x" direction w/o a "y" component.

Any help appreciated.
 

Answers and Replies

  • #2
782
1
I think your formula should be [itex]F_2 = \sqrt{(ma)^2 - F_1^2}[/itex] (remember [tex]c^2 = a^2 + b^2[/tex]) to get 14.09 N.
 
  • #3
61
0
cscott said:
I think your formula should be [itex]F_2 = \sqrt{(ma)^2 - F_1^2}[/itex] (remember [tex]c^2 = a^2 + b^2[/tex]) to get 14.09 N.
This is not the answer... tried it and got it wrong :(

I also tried splitting up the the acceleration into x and y components and trying to solve it F1x+F2x = MAx; F1y+F2y=MAy F1x=0 F1y=12.6N

I solved it and then did sqr rt(F2x^2+F2y^2)


= 16.2662N


I only have 1 try left so it would be great if someone can help me out, please.
 
  • #4
1,860
0
How come the force acting north has a magnitude of 12.6N? It should be [tex]masin(\theta)=F_N[/tex] which is 8.87N. Was 12.6 given?
 
  • #5
Little_Rascal
The force you're trying to find must also have an y-component, otherwise cscotts formula would've worked.
 
  • #6
61
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Mindscrape said:
How come the force acting north has a magnitude of 12.6N? It should be [tex]masin(\theta)=F_N[/tex] which is 8.87N. Was 12.6 given?
The question states (as found in the original post):

"One of the two forces acting on the mass has a magnitude of 12.6 N and is directed north."

So assume the sole direction of F1 is North, or in the positive y direction.

Any ideas as to an equation or problem-solving technique to solve this?
 
  • #7
1,860
0
To make sure I have this right:

-The net force (magnitude) of the two components is the [tex]7 \frac{m}{s^2} * 2.7kg = 18.9N[/tex]

-This magnitude has an angle of 28º north of east

-The north acting component of the net force is 12.6N

Unless physics isn't allowed to use vectors anymore, this is wrong.
 
  • #8
61
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Mindscrape said:
To make sure I have this right:

-The net force (magnitude) of the two components is the [tex]7 \frac{m}{s^2} * 2.7kg = 18.9N[/tex]

-This magnitude has an angle of 28º north of east

-The north acting component of the net force is 12.6N

Unless physics isn't allowed to use vectors anymore, this is wrong.
hmm... i don't know. I cut and pasted the question -- maybe the question is wrong?
 
  • #9
mukundpa
Homework Helper
524
3
ma cos28 = (F2) cos@
ma sin28 = (F1) + (F2) sin@

@ is the angle F2 makes north of east

eliminate @ to get F2

M.P.
 
  • #10
61
0
mukundpa said:
ma cos28 = (F2) cos@
ma sin28 = (F1) + (F2) sin@

@ is the angle F2 makes north of east

eliminate @ to get F2

M.P.

ma cos28 / cos@ = ma sin28 - F1 / sin@

tan@ = ma sin28 - F1/ ma cos28

@ = tan -1 (ma sin28 - F1/ ma cos28)

F2x = 17.09883283
F2y = -17.09883283

F2 = sqr rt ( 17.09883283^2 * 2)

= 24.18140129 N?

This comes up as wrong. I dont know -- maybe I made a math mistake.

Thanks for the help though.
 
  • #11
mukundpa
Homework Helper
524
3
solve in this way

F2 cos@ = macos28 = 16.688N
F2 sin@ = masin28 - F1 = -3.727 N

squaring and adding F2 = 17.1 N (please check the calculations)

Who told you that the two forces are perpendicular or F2 is towards east?

M.P.
 
  • #12
61
0
mukundpa said:
solve in this way

F2 cos@ = macos28 = 16.688N
F2 sin@ = masin28 - F1 = -3.727 N

squaring and adding F2 = 17.1 N (please check the calculations)

Who told you that the two forces are perpendicular or F2 is towards east?

M.P.

I think that's it. Thanks for the help!
 

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