CrazyEngineers Archive
Old, but evergreen and popular discussions on CrazyEngineers, presented to you in read-only mode.
@ErAnushka • 16 Oct, 2012
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 😔
@Jeffrey Arulraj • 17 Oct, 2012 • 1 like larger force is given to P as it causes the most impact in any point acted by two forces
@ErAnushka • 17 Oct, 2012
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 • 17 Oct, 2012 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 • 17 Oct, 2012
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 • 17 Oct, 2012 • 1 like 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 • 17 Oct, 2012 • 1 like
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 • 17 Oct, 2012 jeffrey samuel Saiwal

Thanks but still I'm confused 😕

Please check this screenshot (sorry for the bad quality, I captured without light):
[​IMG]
@ErAnushka • 17 Oct, 2012
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 • 17 Oct, 2012 • 1 like
ErAnushka
jeffrey samuel Saiwal

Thanks but still I'm confused 😕

Please check this screenshot (sorry for the bad quality, I captured without light):
[​IMG]
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 • 17 Oct, 2012
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 • 18 Oct, 2012
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
[​IMG]

👍
10.3k views

Related Posts

@richa22 · Jun 26, 2013

i am doing m.tech in power electronics and systems and i want to make a solar inverter but i heared research has already been done on this project hence kindly...
6.2k views

@vivek kumae · Jan 2, 2013

got 750 rank & opted for DAC course.which institute is best. CDAC bangalore or CDAC MUMBAI(kharghar). whats about VITA, Mumbai (Juhu).plz help me out ..!!
11.9k views

@Kaustubh Katdare · Mar 30, 2015

Chinese engineers seem to be getting their new mega-project ideas from Isaac Asimov's books. According to ET, China is planning to build a mega solar power station at an altitude...
4.1k views

@suyash · Jun 24, 2007

[FONT="]TATVA’07: A RETROSPECTIVE[/FONT] [FONT="]The Annual Technical Festival of the Department of Metallurgical and Materials Engineering, Visvesvaraya National Institute of Technology, Nagpur, had a successful inaugural edition which was organized from...
3.5k views

@Nivedita Gautam · May 19, 2014

I m b.tech computer science student. I am about to complete my 3rd year. I am supposed to do 45 days summer internship. I want to do it in java...
4.1k views