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

[Benjamin_harsh]

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

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.

 

[Wrichik Basu]

The formula (which you have mentioned in the subject line) comes from the Law of Vector Addition. A simple derivation is given here:
https://www.mathstopia.net/vectors/triangle-law-vector-additionYou can look up any HS book. A chapter on Vectors will surely have that derivation.

 

[Benjamin_harsh]

The formula ... have that derivation.

My Question is why $3.cos 60^0$ added to 4?

 

[mjc123]

Because the x component of R is 4 + 3cos60°. Seethe diagram in the link posted by @Wrichik Basu.

 

[Gaussian97]

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.

 

[Wrichik Basu]

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.

 

[Benjamin_harsh]

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.

 

[Benjamin_harsh]

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?

 

[NascentOxygen]

$tan^{–1}(n)$ is read as "the angle whose tan is n". So it's the way you calculate $\theta$ when an equation gives the value for $tan\ \theta$.

 

[ZapperZ]

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.

 

[gmax137]

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.

 

[Benjamin_harsh]

So is my diagram in the post wrong?

 

[Wrichik Basu]

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.

 

[Benjamin_harsh]

I think this is the correct diagram:

But my doubt position of 4N and 3N. Why 4N need to be in base (or) why not 3N need to be in base?

 

[Gaussian97]

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?

It's not important at all.

 

[jbriggs444]

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.

 

[Wrichik Basu]

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?

 

[Benjamin_harsh]

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.

 

[ehild]

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.

 

[BvU]

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 :

In other words: vector addition is commutative

This is what @Gaussian97 means when he/she writes 'It's not important'.

 

[Benjamin_harsh]

if you place the 3 'in base' and add the 4 you get the same resultant R :

So are you saying $\frac{4. sin60^0}{3 + 4.cos 60^0} = \frac{3. sin60^0}{4 + 3.cos 60^0}$ ?

 

[BvU]

No

 

[Benjamin_harsh]

No

How to find angle easily? I am getting confused.

 

[Benjamin_harsh]

$A = 4$ and $B = 3$.

How can you tell $A = 4$ and $B = 3$? Why not vice versa?

 

[BvU]

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

Spoiler

(successively: the right dash-dot line (i.e. opposite), the 3N plus the left dash-dot (adjacent).

 

[Benjamin_harsh]

The first one ... there in my sketch ?

How many diagrams can i draw from the given question?

 

[Wrichik Basu]

How can you tell $A = 4$ and $B = 3$? Why not vice versa?

You can of course change that. Whatever be your choice of A and B, you will get the angle that the resultant makes with either the 4N force, or the 3N force, depending on your choice of A and B.

For example, when you had worked out the problem, you took A=4, so you found the angle between the resultant and 4N. If you took A=3, you will find the angle between the resultant and 3N.

The mod value of the resultant remains invariant irrespective of A and B.

 

[Benjamin_harsh]

For example, when you had worked out the problem, you took A=4, so you found the angle between the resultant and 4N. If you took A=3, you will find the angle between the resultant and 3N.

What formula should i use for finding the angle between the resultant and 4N?

 

[Wrichik Basu]

What formula should i use for finding the angle between the resultant and 4N?

This:

The formula is $$\phi = \tan^{-1}\left( \frac{B \sin \theta}{A+B\cos \theta}\right)$$

$\phi$ is the angle that you marked as $q$. Basically it's the angle between $A$ and the resultant.

You originally took $\theta = 60°$, $A = 4$ and $B = 3$, so you got the angle between 4N and the resultant. If you interchange A and B, you get the angle between 3N and the resultant.