Finding Resultant Force: Calculation and Explanation

[rahul222]

A person on a parachute has an upwards force of 500N and a downward force of 1000N. The wind strikes the person at 45 degrees and has a force of 300N find the resultant force to 1 decimal place.

Hint: Resolve the force of the wind into its horizontal and vertical components, before trying to add the forces together. Remember you can’t add forces that are perpendicular, you’ll have to use Pythagoras!


Plz help me I didn't understand the hint and please show the working of the problem I will be really grateful to you. By the way someone told me the answer is 357.6 N but he doesn't know the working so please help and reply soon. :)

 

[Redbelly98]

Welcome to PF.

Start by drawing a force diagram for the person. There are 3 forces here.

p.s. You didn't say whether the wind force is 45° upward or downward from horizontal. Is that specified in the problem statement?

 

[rahul222]

Sorry forgot to specify that it is upwards actually i have the diagram but i don't know how to post it?

 

[Nabeshin]

Just host it on photobucket or some other image hosting server and post the link. Attaching files requires them to be reviewed, which takes forever.

 

[rahul222]

sorry but that will take time and i want the answer qucikly sorry. Plz people help me find the answer

 

[rahul222]

Plz explain the answers also so that I can perform better and I have a test soon so ur info will be invauable to me thank you reply soon

 

[Redbelly98]

Have you drawn a force diagram yet, as I suggested in post #2?

Don't worry about posting the diagram here, just draw it for yourself.

 

[rahul222]

i drew it but colud u please say the steps

 

[#matt#]

Seems like you can just determine the delta of your upward force(1000-500) and use it as the first leg of your diagram triangle, create a second leg with a 45 degree angle to the initial leg, and give it the wind's force of 300.

Once you draw this, you should be able to determine the value of the third leg using the laws of cosine.

Hope that helps!

 

[Redbelly98]

To find the net force, the steps are:

Draw a force diagram, and label all of the forces.

Decide what the positive-x direction is to be.

Decide what the positive-y direction is to be.

Figure out the x- and y-components of each force. You can either use trig, or properties of 45-degree right triangles.

Add up the x-components of each force to get the total x-component.

Add up the y-components of each force to get the total y-component.

Use the Pythagorean Theorem to find the magnitude of the force.

Use trig to find the direction of the force.

 

[rahul222]

didn't understand redbelly98 please could u explain in detail by doing it i will be grateful to u.

 

[Redbelly98]

No, I can't do your homework for you ... it's forum policy.

 

[rahul222]

redbelly98 this is not my homework i m new to this forum and actually i m goin a bit ahead and doing more work than my teacher gave me and this question is taken from a website so please help me solve this question and i guarantee u this is not a homework question so please help me solve it if u don't believe i can even give u the site where i got the question from so please help me redbelly98

 

[Redbelly98]

I can help but you will have to do some work as well. That's how this forum works, and that's how I work.

You said you have drawn a force diagram. Did you label the 3 forces?

Then the next step is to choose a direction for +x and +y. Can you do that?

If you can find a way to post your diagram -- with forces labelled, and +x and +y directions indicated -- we could proceed to what's next.

If you are confused by labeling the forces and +x and +y, then post what you have drawn so far.

 

[rahul222]

I can post the question

http://tinypic.com/view.php?pic=200eljc&s=4

here it is

 

[rahul222]

I will now post my force diagram which I have drawn,

http://tinypic.com/view.php?pic=2qscmqr&s=4


I am not sure if I am going correct and I don't understand labelling the force please help redbelly 98.

 

[rahul222]

Oops I forgot to put the arrows the 300 N is going upwards North East direction and 500 N downwards

 

[rahul222]

please help me finish the question soon redbelly98

 

[Redbelly98]

There is a problem with your force diagram. The 300 N force is at a downward angle, but you said earlier it is at an upward angle. See if you can fix that.

If the drawing was made in Word or Excel, you could put arrow heads on the lines to indicate directions more clearly.

p.s. please stop rushing things. I get the message, you're in a hurry ... even though you have gone ahead of the class (?). I'm busy and am giving you what time I can.

 

[rahul222]

Sorry redbelly98 for getting u angry i was being too selfish and I forgot even u have work.

I didn't get wht u meant by "The 300 N force is at a downward angle".Isn't the force goin upward?. So I can't amend the force diagram. So please could u make tht point clear and then maybe i can change the force diagram. Also please tell what is the mistake i did in the force diagram so tht i can make changes. Thanku

 

[Redbelly98]

New approach: try the following:

Go to the website below, and enter the following numbers:
Length of red vector: 3
Length of blue vector: 5
Angle of red vector: 45
Angle of blue vector: 270

The red vector would be the 45 degree wind force in your question. The blue vector would be the result of the 2 forces that act upward and downward. The green vector will be the final resultant force.

Here is the website:
http://physics.bu.edu/~duffy/java/VectorAdd.html

 

[rahul222]

I did wht u said. Now...

 

[rahul222]

I actually understand something but why do we take 270 degrees for blue shouldn't it be 180 as it is downwards??

 

[Redbelly98]

Look at the blue vector in the diagram. Which way does it point when you use 270? Which way when you use 180?

 

[rahul222]

K redbelly98 wht next?? And was it only in this case or we always measure from horizontal.

 

[Redbelly98]

Not always, but in many cases yes, angles are measured from the horizontal.

Next is to decide which directions are to be positive. Let's take to-the-right as positive in the x-direction, and upward as positive in the y-direction.

After that, you need to find the x- and y-components of the two Force vectors.

If you haven't done that before, this links shows how:
http://www.1728.com/vectutor.htm

 

[rahul222]

I finished doing it and thank you redbelly98 u were of great help and I hope u conitinue helping me in future. Do u have any email address??

 

[Redbelly98]

Your welcome.

My personal policy is that contact with people in this Physics Forum stays in the forum, and to use email just for people I know in "real life". It's nothing personal, but I've had problems with other people in past experiences.

Good luck with the rest of your studies.

 

[rahul222]

Its k
but just one last question when i use the cos rule which is square root
a^2+b^2-2bcCos A i get the magnitiude correct 3.58 with it but why when i use the sine rule to find the angle i get something 81?

 

[Redbelly98]

What angle is the sine rule saying is 81°?

 

[rahul222]

angle between the resultant and the wind (on the left side of resultant). If u go in the vector adddition site which u gave me the first one http://physics.bu.edu/~duffy/java/VectorAdd.html
in this site the angle between green and red in the question.Hopefully u got it and can explain me.

 

[Redbelly98]

This sounds reasonable, that the resultant and wind vectors are 81 degrees apart.