Statics: Magnitudes/Resultant force

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The discussion focuses on solving a statics problem involving resultant forces and their angles. Participants calculate the components of two forces, F1 and F2, and express uncertainty about the angles and resultant calculations. They emphasize the importance of consistent sign usage in summing the forces and clarify that the angle should be measured counterclockwise from the positive x-axis. The final calculations yield a resultant force of approximately 747N at an angle of 85.5 degrees, with suggestions to visualize the vectors for better understanding. The conversation highlights the need for clarity in interpreting diagrams and applying trigonometric functions correctly.
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Homework Statement


I'll attatch it.

The Attempt at a Solution


Alright, so I work out the angle of the right side...

cos-1(12/13) = 22.6°

Find the components of the right side...

F1x = 800cos(40)
= 612.84
F1y = 800sin(40)
= 514.23

F2x = 600cos(22.6)
= 553.92
F2y = 600sin(22.6)
= 230.58

I'm not sure where to go from here, not even sure if the angles right. Looking forward to some help thanks.
 

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Add them up remembering to be consistent with the sign (especially for Total Fx).

eg Total Fy = F1y+F2y

Then check the answer is in the format requested.
 
Total Fy = 744.81
Total Fx = 58.92

Cause one of the x's is going in the opposite direction right?

But the answers meant to be:
F = 747N
Angle = 85.5degrees
 
Reconstruct FTotal using pythagorous..

FTotal = SQRT (Fy2+Fx2)

likewise the angle
 
FTotal = 747N

but with the angle what would I need to use? Sine rule or something?
 
Still struggling to find out how to find the angle of the triangle.
 
Soh - sin (theta) = opposite/hypotenuse
Cah - cos (theta) = adjacent/hypotenuse
Toa - tan (theta) = opposite/adjacent
 
dimens said:
Still struggling to find out how to find the angle of the triangle.

You know the length of all three sides. Draw it on your diagram. Apply basic trig. eg
Angle = Cos-1(... you know the rest.
 
the diagram isn't a right angled triangle though? how can i use them trig function?
 
  • #10
I got the answer... from cos^-1(58.92/744.8)
= 85.46
= 85.5

But I'm struggling to get my head around the diagram. Where's the angle that I'm meant to be finding and what does it mean by "it's direction compared to the x-axis clockwise."
 
  • #11
You chose +ve to be to the the right and since Total Fx is positive the Total vector points to the right of vertical.

Total Fy = 744.81
Total Fx = 58.92

Since Total Fy >> Total Fx the resulting vector is almost vertical.

So by inspection the resulting vector is almost but not quite vertical and points/leans slightly to the right. You calculated about 85.5 degrees to the horizontal which confirms this. Draw it on the diagram.

The question asked for the angle relative to the x-axis anticlockwise. Presumably they mean the anti clockwise direction is to be considered positive. So this also 85.5 degree angle you calculated.
 
  • #12
Tell you what, why not draw vectors representing

Total Fy
Total Fx
and the resulting vector Total F

on the diagram and post it again.
 
  • #13
Hi guys I'm back with another one, this ones doing my head in I can't seem to solve it...

QUESTION:
Determine the magnitude of resultant force and its direction, measured counterclock wise from the positive x axis; If Fa= 770N, Fb= 240N and Theta= 35 degrees.

ATTEMPT AT PROBLEM:
So I get the components...

Fax = 770*cos(36.87) = 616
Fay = 770*sin(36.87) = 462
Fbx = 240*cos(35) = 196.6
Fby = 240*sin(35) = 136.7

Fx = 812.6
Fy = 592.7

sqrt(812.6^2 + 592.7^2) = 100.9.336N

to get angle:

tan-1(812.6/592.7) = 53.9°

Answers meant to be:
R= 732 N
Theta= 55.04 degrees
 
  • #14
noticed that the picture didn't attach, here it is..

418jr.jpg
 

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