February 26th, 2018, 09:13 AM  #1 
Newbie Joined: Feb 2018 From: England Posts: 12 Thanks: 0  Vectors / Magnitude
Hi All, Having great difficulty in the following problem: Magnitude of the 3 forces acing on a stable object A = 800kN West B = 650kN South C = ? What is the magnitude of force C? Any help would be much appreciated. 
February 26th, 2018, 10:01 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,961 Thanks: 1605 
Are you required to use trigonometry rather than geometry?

February 26th, 2018, 10:27 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,163 Thanks: 867 
I would write the vectors in "components", taking the positive xaxis to be "east" and the positive yaxis to be "north" A is "800 kN west" so A= <800, 0>. B is "650 kN south" so B= <0, 650>. The sum of those is <800, 650> so, in order that the object not move, C must be <800, 650>. The magnitude of C is $\displaystyle \sqrt{800^2+650^2}$. 
February 27th, 2018, 01:08 AM  #4 
Newbie Joined: Feb 2018 From: England Posts: 12 Thanks: 0 
Thanks for the replies, much appreciated. Skipjack  With a limited question I am not sure however, if there are two ways I can answer the question I'll do both if you could shed any light? Country boy  Thanks for the detailed response. Makes sense now you've explained, so the magnitude of C = 1030.8kN? 1) How would I go about calculating the angles of each? 2) If the object has to be moved directly south. In order to achieve this what would have to happen to the resultant force acting on the object? (I assume I would have to calculate the resultant force first?) 3) State two ways the individual forces could be changed to achieve this resultant force? May change the magnitude and direction of any of the individual forces. Thanks for your time and support. 
February 27th, 2018, 02:52 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 18,961 Thanks: 1605  Forces.PNG The forces can be "chained", as in the above diagram. If the resultant force is nonzero, the corresponding diagram would show a fourth line (for the resultant). It should be obvious that trigonometry could have been used instead of Pythagoras. Use trigonometry to calculate the angles. It's usually convenient to start by using coordinate geometry. If you need to use trigonometry, a fullyannotated diagram is recommended. 
February 27th, 2018, 04:25 AM  #6 
Newbie Joined: Feb 2018 From: England Posts: 12 Thanks: 0 
Thanks again skipjack appreciate your help. Correct me if I'm wrong, with a resultant force of zero? What would happen to the resultant force once the object is moved directly South. Could this be easily achieved my changing the magnitude and direction of the other individual forces? 
February 28th, 2018, 12:26 AM  #7 
Newbie Joined: Feb 2018 From: England Posts: 12 Thanks: 0 
Any other feedback on this would be much appreciated. Correct me if I'm wrong, with a resultant force of zero? What would happen to the resultant force once the object is moved directly South. Could this be easily achieved my changing the magnitude and direction of the other individual forces? Thankyou All. 
February 28th, 2018, 12:35 AM  #8  
Senior Member Joined: Jun 2015 From: England Posts: 823 Thanks: 243  Quote:
A little tip. (useful theorem) When a body is in equilibrium (your stable) under the action of three forces in the same plane, those three forces meet at a point.  
February 28th, 2018, 06:17 AM  #9 
Newbie Joined: Feb 2018 From: England Posts: 12 Thanks: 0 
Thanks studio. Is there a way to express the resultant force? My theory is (correct me if wrong please).... In order to move the object directly south: 1) The object is no longer in equilibrium, the resultant force would have to become dynamic? 2) Two individual forces that could achieve this: Force B Magnitude increased? Direction of Force A adjusted to that similar of Force B? Many thanks. 
February 28th, 2018, 06:37 AM  #10  
Senior Member Joined: Jun 2015 From: England Posts: 823 Thanks: 243  Quote:
That is the resultant = 0. This allows us to write the equations of equilibrium, perhaps in component form as country boy suggests. Alternatively you can note that the polygon ( a triangle in this case since there are 3 forces is closed. If the three forces are not in equlibrium then the polygon is open. Open means that when you place one vector after another, head to tail, in the manner showed by skipjack, you do not end up back where you started. The end point is not the same as the start point. You need to choose one method, have you been taught one? and did you understand my comment about 'stable' meaning that the object is in equilibrium? Last edited by studiot; February 28th, 2018 at 06:40 AM. Reason: spelling  

Tags 
magnitude, vectors 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
3D Vectors (Dot product, magnitude, and angles)  jaredbeach  Calculus  1  January 16th, 2013 09:37 AM 
Vector magnitude  LordRaaa  Algebra  3  May 24th, 2012 12:02 PM 
finding vectors of magnitude  LordRaaa  Linear Algebra  1  May 17th, 2012 12:49 PM 
What is the magnitude of v  remeday86  Calculus  3  March 10th, 2009 03:55 PM 
magnitude  remeday86  Calculus  3  March 8th, 2009 05:23 PM 