Problem: Law of Parallelogram of Forces [Help me]

Replies

• 

  Jeffrey Arulraj

  larger force is given to P as it causes the most impact in any point acted by two forces
• 

  ErAnushka

  jeffrey samuel

  larger force is given to P as it causes the most impact in any point acted by two forces

  Thank you 😀 I'll take larger force as "P" 😀
  
  One more question, the formula for magnitude of resultant R is:
  
  R^2 = P^2 + Q^2 + 2PQ cos @
  
  
  But what if "Q" = "P" ?
  in the book, I can see that if P = Q then R^2 = P^2 + P^2 - 2P^2 cos @
  
  I didn't get it, what is the reason that makes the 2P^2 negative 😕
• 

  Jeffrey Arulraj

  What is the angle between the two forces in concern it does not arbitrarily become negative do refer other sources also
  
  Kindly clarify me in this about the angle between the two forces if it is greater than 90 then the negative sign will come
• 

  ErAnushka

  jeffrey samuel

  What is the angle between the two forces in concern it does not arbitrarily become negative do refer other sources also
  
  Kindly clarify me in this about the angle between the two forces if it is greater than 90 then the negative sign will come

  
  Here is the problem: Two forces act at an angle of 60, their resultant is 50 N acting at 30 with one of the forces, find the value of forces.
  
  
  It's little complicated, but in the end we get P = Q, and that formula of magnitude changed 😕 I hope you've solution 😀
• 

  Jeffrey Arulraj

  I have very less touch in these problems now but I can try it out if you tell what is 30 here bit confused in it 😔
• 

  Saiwal

  ErAnushka

  Thank you 😀 I'll take larger force as "P" 😀
  
  One more question, the formula for magnitude of resultant R is:
  
  R^2 = P^2 + Q^2 + 2PQ cos @
  
  
  But what if "Q" = "P" ?
  in the book, I can see that if P = Q then R^2 = P^2 + P^2 - 2P^2 cos @
  
  I didn't get it, what is the reason that makes the 2P^2 negative 😕

  Its a printing mistake the sign is always positive no matter what the angle.
  if the angle is 180 and both forces are equal the resultant comes out zero.
  

  ErAnushka

  I'm solving some problems based on law of parallelogram of forces. I completely understand the concept and how to prove that 😀 But I've one silly question, look at problem and please suggest me, how to choose the forces as P and Q?
  
  problem: Given two forces of magnitude 10kN and 20kN, are having a resultant of 15kN. Find the angle between two forces and the direction of the resultant 😀
  
  We know the formula:
  for resultant direction:
  
  A (alpha) = tan^-1 ((P sin @) / (P cos @ + Q))
  
  
  But if we chose wrong "P" and "Q" then we won't get the correct answer!
  
  
  My question is: In the book, the writer assigned 20kN to "P" and 10kN to "Q"
  
  But if I take P = 10kN and Q = 20kN then the answer will be different,
  
  How do I decide which is "P" force and which one is "Q" force? 😀
  
  Please help 😔

  The resultant makes angle with both 'P' and 'Q'.
  If you choose 20kN as 'P', you'll get the angle between 20kN force and resultant 'R' (alpha).
  If you choose 10kN as 'P', you'll get angle between 10kN force and the resultant which will be equal to (@-alpha).
  Hope this helps.
• 

  ErAnushka

  #-Link-Snipped-# #-Link-Snipped-#
  
  Thanks but still I'm confused 😕
  
  Please check this screenshot (sorry for the bad quality, I captured without light):
  
• 

  ErAnushka

  Saiwal

  The resultant makes angle with both 'P' and 'Q'.
  If you choose 20kN as 'P', you'll get the angle between 20kN force and resultant 'R' (alpha).
  If you choose 10kN as 'P', you'll get angle between 10kN force and the resultant which will be equal to (@-alpha).
  Hope this helps.

  
  
  @ - Alpha, doesn't work 😔
  
  You can try it:
  
  P = 20kN,
  Q = 10kN,
  (Resultant) R = 15kN,
  
  Formula to find A (alpha) = tan^-1 ((P sin @) / (P cos @ + Q))
  
  You can also check in screenshot 😀
  
  
  
  Now the answer will be -75.53, but we need positive angle with respect to x-axis so we take 180 - 75.53 = 104.48 😀
  
  
  
  Now, if we take :
  P = 10kN,
  Q = 20kN,
  (Resultant) R = 15kN,
  
  Formula to find A (alpha) = tan^-1 ((P sin @) / (P cos @ + Q))
  
  The answer will be different! I got 49.28 😨
  
  as per your instructions: 180 - 49.28 = 130.72 😔
  
  I think the larger force is assigned to P and less force is assigned to Q, not sure though. 😳
• 

  Saiwal

  ErAnushka

  #-Link-Snipped-# #-Link-Snipped-#
  
  Thanks but still I'm confused 😕
  
  Please check this screenshot (sorry for the bad quality, I captured without light):
  

  the value of P is 28.868kN
  there is no way that two forces at 60 deg result in another 50kN resultant. that's just plain wrong.
  will post answer to your other question shortly 😀
• 

  ErAnushka

  Saiwal

  the value of P is 28.868kN
  there is no way that two forces at 60 deg result in another 50kN resultant. that's just plain wrong.
  will post answer to your other question shortly 😀

  
  Okay, and I got the same answer ^_^
• 

  Saiwal

  ErAnushka

  @ - Alpha, doesn't work 😔
  
  You can try it:
  
  P = 20kN,
  Q = 10kN,
  (Resultant) R = 15kN,
  
  Formula to find A (alpha) = tan^-1 ((P sin @) / (P cos @ + Q))
  
  You can also check in screenshot 😀
  
  
  
  Now the answer will be -75.53, but we need positive angle with respect to x-axis so we take 180 - 75.53 = 104.48 😀
  
  
  
  Now, if we take :
  P = 10kN,
  Q = 20kN,
  (Resultant) R = 15kN,
  
  Formula to find A (alpha) = tan^-1 ((P sin @) / (P cos @ + Q))
  
  The answer will be different! I got 49.28 😨
  
  as per your instructions: 180 - 49.28 = 130.72 😔
  
  I think the larger force is assigned to P and less force is assigned to Q, not sure though. 😳

  here's the solution:
  hope it helps
  
  
  👍