How to find magnitude of two forces

  • #1
11
0
Two forces are given and are applied to a car in an effort to accelerate it: 414N at 9 degrees and 340N at 25 degrees. What is magnitude of the resultant of the two forces? Answer in units of N.

So I've added together X= (414cos9) + (340cos25), and got -40.
For the Y, Y= (414sin9) + (340sin25)=125.
I took square root of (-40)^2 + (125)^2, and got 131.24.

However, that is not the correct answer. What am I doing wrong? Help please?!
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
13,096
3,416
Note the angles are in degrees, not radians.
 
  • #3
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,670
Two forces are given and are applied to a car in an effort to accelerate it: 414N at 9 degrees and 340N at 25 degrees. What is magnitude of the resultant of the two forces? Answer in units of N.

So I've added together X= (414cos9) + (340cos25), and got -40.
You've added two positive quantities together and obtained a negative result. How? o0)

For the Y, Y= (414sin9) + (340sin25)=125.
I took square root of (-40)^2 + (125)^2, and got 131.24.

However, that is not the correct answer. What am I doing wrong? Help please?!
 
  • #4
phinds
Science Advisor
Insights Author
Gold Member
16,921
7,890
Well you are computing the right values to add together for X and also for Y but the answers you get for those computation are just weird.
 
  • #5
11
0
Steamking, 414 x cos 9 degrees = - 377.2.
 
  • #6
TSny
Homework Helper
Gold Member
13,096
3,416
Check your calculator to make sure it is in degree mode.
 
  • #7
11
0
Note the angles are in degrees, not radians.
So what you are saying Tsny is that I should convert the answers to radians?
 
  • #8
TSny
Homework Helper
Gold Member
13,096
3,416
Keep the angles in degrees and just switch the angle mode of your calculator to degrees instead of radians. Your calculator is assuming that your angles are in radians.
 
  • #9
11
0
Keep the angles in degrees and just switch the angle mode of your calculator to degrees instead of radians. Your calculator is assuming that your angles are in radians.
I've switched the mode to degrees, but still am getting the wrong answer. :(
 
  • #10
TSny
Homework Helper
Gold Member
13,096
3,416
what does your calculator now give you for cosine of 9 degrees?
 
  • #11
11
0
what does your calculator now give you for cosine of 9 degrees?
.9876883
 
  • #12
TSny
Homework Helper
Gold Member
13,096
3,416
OK good. Was a diagram included in the problem? Are the two angles measured from the x axis? Are both angle above the x axis, or is one angle above the axis and the other angle below the axis?
 
  • #13
11
0
OK good. Was a diagram included in the problem? Are the two angles measured from the x axis? Are both angle above the x axis, or is one angle above the axis and the other angle below the axis?
340 N @ 25 degrees is below x-axis. 414 N @ 9 degrees I above x-axis.
 
  • #14
TSny
Homework Helper
Gold Member
13,096
3,416
Are the Y components of both forces positive? If not, which force has a negative Y component?
 
  • #15
11
0
Yes
Are the Y components of both forces positive? If not, which force has a negative Y component?
 
  • #16
TSny
Homework Helper
Gold Member
13,096
3,416
The 340 N points below the x axis. Draw this force and draw its Y component. Does the Y component point up or down?
 
  • #17
11
0
I
The 340 N points below the x axis. Draw this force and draw its Y component. Does the Y component point up or down?
It is pointing up
 
  • #19
11
0
Ye
Hmm. I think you need to review the concept of components of a vector. Particularly their signs.

For example, take a look at the picture at the bottom of this link: http://www.texttutoring.com/an-arrow-is-shot-off-a-cliff-at-an-angle-with-crosswind/

Does it make sense to you that the Y component of the vector is negative (pointing down)?

I think once I get a good look at this tomorrow, it'll make better sense. As far as I can see, the vector does move in a negative direction towards F2.
 
  • #20
11
0
Ye


I think once I get a good look at this tomorrow, it'll make better sense. As far as I can see, the vector does move in a negative direction towards F2.
Thanks for all your help TSny!
 
  • #21
TSny
Homework Helper
Gold Member
13,096
3,416
OK, good luck with it. I'll just leave you with this.

In the picture below, the red vector has both positive X and Y components. (If you walked from the tail to the head of the vector you would be going in the positive X and positive Y direction.) But the blue vector has a positive X component and a negative Y component. (If you walked from the tail to the head of this vector you would be going in the positive X but in the negative Y direction.)
 

Attachments

  • vectors.png
    vectors.png
    1.6 KB · Views: 350
  • #22
11
0
OK, good luck with it. I'll just leave you with this.

In the picture below, the red vector has both positive X and Y components. (If you walked from the tail to the head of the vector you would be going in the positive X and positive Y direction.) But the blue vector has a positive X component and a negative Y component. (If you walked from the tail to the head of this vector you would be going in the positive X but in the negative Y direction.)

Yes, thank you for this! I had actually gone off of a sketch I had, and it was not accurate. After you had mentioned so, I looked back at the hw and it's sketched the way you have it. Thanks for all your explanations!
 

Related Threads on How to find magnitude of two forces

  • Last Post
Replies
2
Views
22K
Replies
2
Views
2K
Replies
1
Views
3K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
9
Views
8K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
21K
Top