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# Work out the resulant force

Asked Sep 21, 2009, 11:01 AM — 7 Answers
I have to figure out the resultant force of this system, and I haven't a clue where to start :(. How do I start to solve this question, can someone please explain to me how to go about tackling this.

I have added an attachment picture which I created in paint with the diagram.

I have literally just signed up to this site minutes ago in desperation as this question has my head scrambled!

Thank you for any help received!

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7 Answers
 ebaines Posts: 11,888, Reputation: 1274 Expert #2 Sep 21, 2009, 11:20 AM

Here's how to solve this. You have three forces acting on the point. Start by resolving each of the three into horizontal and vertical components. The horizontal component of a force $F$ is $F_x =F*cos(angle)$, and the vertical component is $F_y = F *sin(angle)$. Then you can add together the three horizontal components to get the total horizontal force $F_{Total_x}$and the three vertical components to get the total vertical force $F_{Total_y}$. Finally, combine the total horizontal and total vertical components into a single vector. The magnitude of the final total force vector is:

$
|F| = \sqrt {F_{Total_x} ^2 + F_{Total_y} ^2 }
$

and the angle of F is

$
arctan(\frac {F_{Total_y}} {F_{Total_x}})
$

Post back with the answer you get.
 RossD1988 Posts: 6, Reputation: 1 New Member #3 Sep 21, 2009, 11:25 AM
Hi, thank you for your quick reply! Though, I am still confused :(.

How do I resolve each of the three into horizontal and vertical components?

I have not done maths/physics in 5 years so its taking time to get used to this again!
 ebaines Posts: 11,888, Reputation: 1274 Expert #4 Sep 21, 2009, 11:40 AM

OK, I'll do F1 for you. The horizontal component of F1 is 15 KN * cos(45) = 15 KN* sqrt(2)/2 = 10.61 KN to the right. The vertical component is 15 KN * sin (45) = 10.61 KN upward.

Now, can you do F2 and F3? For F2 the angle is 0 degrees and for F3 the angle is minus 60 degrees.
 RossD1988 Posts: 6, Reputation: 1 New Member #5 Sep 21, 2009, 03:12 PM
For F3, am I correct in saying it is 15KN* sin(45)=10.60KN

The notes that I was given for this question did not have a worked example which I was hoping for, and it doesn't mention how to work out these questions either.

We were given this question, and the answer (which is 38.2KN + 3.6degrees) and basically let go to work it out.

All I need is 1 good worked example and il b able to work this stuff out for myself :(.
 RossD1988 Posts: 6, Reputation: 1 New Member #6 Sep 22, 2009, 02:01 AM
Ignore that last post of mine, I put in the wrong angle!

So far I have wrote down:

F1 15KN COS(45) = 10.6KN
F2 15KN COS(60) = 7.5
F3 20KN COS(0) = 20
= 38.1 KN

38.1KN is the correct answer that was given to us on the sheet given. We were given the correct answer and the problem but we had to work out how to get the answer.

Though, you were saying to figure out the vertical force also? How do u do this? And this (based on looking at your previous posts) is how to work out the resulant angle?
 ebaines Posts: 11,888, Reputation: 1274 Expert #7 Sep 22, 2009, 05:43 AM

What you have so far is the sum of the forces in the X (horizontal) direction. So far so good - but you are not done! Your answer of 38.1KN does not equal the answer you were given, which is 38.2KN at 3.6 degrees. What you need to do next is add up the forces in the Y (vertical) direction; you need to calculate F1*sin(45) + F2*sin(0) + f3*sin(-60).

Once you have that, calculate the total force and direction using the formulas I gave you earlier in post #2. Hope this helps.
 RossD1988 Posts: 6, Reputation: 1 New Member #8 Sep 23, 2009, 02:33 PM

Thank you very much for your help I have got the answer to that and many other resultant force questions now. :)

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