Why ##3.cos 60^0## is added to 4?

  • #1

Homework Statement:

Two forces of 3 N and 4 N are acting at a point such that the angle between them is 60 degrees. Find the resultant force.

Relevant Equations:

##tan(q) = \frac{3. sin60^0}{4 + 3.cos 60^0}## = 0.472
243998


Magnitude R of the resultant force is R =
##\sqrt {3^2+4^2+2∗3∗4∗cos60^0}##

= ##\sqrt {9+6+12}##

= ##\sqrt 37 = 6.08 N##

Direction of R is given by finding the angle q,

##tan(q) = \frac{3. sin60^0}{4 + 3.cos 60^0}## = 0.472

##q=tan^−1(0.472)q= tan^−1(0.472) = 25.3^0##

Thus R is 6.08N in magnitude and is at an angle of ##25.3^0## to the 4N force.
 

Attachments

Last edited:

Answers and Replies

  • #4
mjc123
Science Advisor
Homework Helper
1,009
478
Because the x component of R is 4 + 3cos60°. Seethe diagram in the link posted by @Wrichik Basu.
 
  • #5
374
174
My Question is why ##3.cos 60^0## added to 4?
Do you know how to compute the components of a vector? If you know that then I suppose that you will also know that the direction of a vector is given by the angle with the ##x##-axis (that's what you are doing) and how it's related with the components. Finally knowing that when summing vectors you have to sum every component the formula for the angle it's almost trivial.
 
  • #6
1,522
1,386
My Question is why ##3.cos 60^0## added to 4?
The formula is $$\phi = \tan^{-1}\left( \frac{B \sin \theta}{A+B\cos \theta}\right)$$ In your case, ##\theta = 60°##, ##A = 4## and ##B = 3##. If you still ask "why", look at the derivation.
 
  • #7
The formula is $$\phi = \tan^{-1}\left( \frac{B \sin \theta}{A+B\cos \theta}\right)$$ In your case, ##\theta = 60°##, ##A = 4## and ##B = 3##. If you still ask "why", look at the derivation.
Now I understand very much. :smile:
 
  • #8
Because the x component of R is 4 + 3.cos60°. Seethe diagram in the link posted by @Wrichik Basu.
In the diagram, base is 4 & opposite is ##3.sin\theta##. I want to use trigonometry ## tan\theta = \frac{opposite}{adjacent}##. But why we need to use another ##tan \theta## i.e ##tan^{-1}(\frac{B \sin \theta}{A+B\cos \theta})##formula here?
 
  • #9
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
##tan^{–1}(n)## is read as "the angle whose tan is n". So it's the way you calculate ##\theta## when an equation gives the value for ##tan\ \theta##.
 
  • #10
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
35,847
4,664
In the diagram, base is 4 & opposite is ##3.sin\theta##. I want to use trigonometry ## tan\theta = \frac{opposite}{adjacent}##. But why we need to use another ##tan \theta## i.e ##tan^{-1}(\frac{B \sin \theta}{A+B\cos \theta})##formula here?
Again, this is a math issue.

Question: Is there a reason why you are stubbornly-refusing to solve problems like this using vector components? I mean, there IS a valid reason for using such a method, such as when you are faced with more than 2 vectors to add (hint: you WILL encounter this when you do E&M in your General Physics class, trust me!). If you are unable do this via vector component, then you are going to have load of "fun" soon enough.

Zz.
 
  • Like
Likes Wrichik Basu
  • #11
1,821
966
My Question is why ##3.cos 60^0## added to 4?
Make a scale drawing using a ruler and protractor. You will see the "end" of R is further to the right than 4. This is because the angle between the 4N and the 3N is not 90 degrees.
 
  • #12
So is my diagram in the post wrong?
 
  • #13
1,522
1,386
So is my diagram in the post wrong?
It's not wrong because your diagram is not to scale. If, however, you draw a very accurate diagram, measuring every length and angle properly, it will deviate from what you've drawn. But most of the time (say about 98% of the time), such precise diagrams are not required if you solve the problem analytically. An accurate drawing, however, will help you visualise the robben better.

Beginners are advised to accurately draw diagrams at first. Then, after one point of time, when you can visualise the situation, you can do the work using rough drawings. Even diagrams can be skipped if you are confident enough that you can do without them.
 
  • Like
Likes phinds
  • #14
I think this is the correct diagram:

244053


But my doubt position of 4N and 3N. Why 4N need to be in base (or) why not 3N need to be in base?
 
Last edited:
  • #16
jbriggs444
Science Advisor
Homework Helper
2019 Award
9,202
3,902
I think this is the correct diagram:

View attachment 244053

But my doubt position of 4N and 3N. Why 4N need to be in base (or) why not 3N need o be in base?
There is nothing preventing you from labelling the 3N side as the base of the triangle and drawing the triangle so that side is horizontal. That would put you in position to conveniently calculate the angle between the "3N" and the "R" side. You could then use that information to determine the angle between the "4N" and the "R" side.
 
  • #17
1,522
1,386
But my doubt position of 4N and 3N. Why 4N need to be in base (or) why not 3N need to be in base?
Do an exercise: take 3N as the base of the triangle. Now calculate the value of the resultant force, and the angle it makes with the 4N force. What do you find?
 
  • #18
I think
Do an exercise: take 3N as the base of the triangle. Now calculate the value of the resultant force, and the angle it makes with the 4N force. What do you find?
I think ##tan \theta = \frac {4}{3}## Since base is 3.
 
  • #19
ehild
Homework Helper
15,526
1,901
I think this is the correct diagram:

View attachment 244053

But my doubt position of 4N and 3N. Why 4N need to be in base (or) why not 3N need to be in base?
You cut the left (red) triangle form the original parallelogram and move it to the other side: You get the blue right triangle with hypotenuse R and legs 3cos(60 ) and 4+3sin(60), and angle q. θ=60, the angle between the 3N and 4N forces.

244054
 
  • #20
BvU
Science Advisor
Homework Helper
2019 Award
13,597
3,285
I think this is the correct diagram:

View attachment 244053

But my doubt position of 4N and 3N. Why 4N need to be in base (or) why not 3N need to be in base?
No need for doubt: if you place the 3 'in base' and add the 4 you get the same resultant R :

244056


In other words: vector addition is commutative

This is what @Gaussian97 means when he/she writes 'It's not important'.
 
  • #21
if you place the 3 'in base' and add the 4 you get the same resultant R :
So are you saying ##\frac{4. sin60^0}{3 + 4.cos 60^0} = \frac{3. sin60^0}{4 + 3.cos 60^0}## ?
 
  • #22
BvU
Science Advisor
Homework Helper
2019 Award
13,597
3,285
No
 
  • #25
BvU
Science Advisor
Homework Helper
2019 Award
13,597
3,285
So are you saying ##\frac{4. sin60^0}{3 + 4.cos 60^0} = \frac{3. sin60^0}{4 + 3.cos 60^0}## ?
The first one is the tangent of the angle between R and the 3N
The second one is the tangent of the angle between R and the 4N
This last one has been chewed out in this thread already.

For the ##\frac{4. sin60^0}{3 + 4.cos 60^0} ## do you see the line segments sitting there in my sketch ?

Don't click before you have your own answer
(successively: the right dash-dot line (i.e. opposite), the 3N plus the left dash-dot (adjacent).
 

Related Threads on Why ##3.cos 60^0## is added to 4?

  • Last Post
Replies
2
Views
5K
Replies
3
Views
4K
Replies
4
Views
1K
Replies
2
Views
6K
Replies
7
Views
8K
Replies
3
Views
7K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
971
Replies
3
Views
6K
Top