Adding and subtracting vectors using vector diagrams

  • Thread starter Thread starter Ishfa
  • Start date Start date
Ishfa
Messages
3
Reaction score
0
Homework Statement
Classical Mechanics
Relevant Equations
s= (a^2 + b^2 + 2ab cos alpha)^1/2
Screenshot 2025-09-25 171406.webp
Screenshot 2025-09-25 171421.webp
 
Physics news on Phys.org
:welcome:

Look at the picture:
Doesn't it seem strange that

1758800807980.webp


Check your math -- and is your calculator set up for radians or for degrees ?

##\ ##
 
Ishfa said:
Relevant Equations: s= (a^2 + b^2 + 2ab cos alpha)^1/2

View attachment 365813View attachment 365814
It would be best to sort out part a) first. There are some mistakes. The first two are:

1) The diagram is wrong. When adding vectors using a diagram, you draw the vectors 'tip-to-tail' or construct a suitable parallelogram.

2) The correct formula for the cosine law is ##c^2 = a^2 + b^2~–~ 2ab \cos\alpha## (where ##\alpha## is the internal angle between sides a and b). But you have used a plus sign.

Surprisingly I agree with ##|\vec a + \vec b| = 4.4587##. It looks like some mistakes cancelled!

Minor edit.
 
Last edited:
Steve4Physics said:
Surprisingly I agree with ##|\vec a + \vec b| = 4.4587##. It looks like some mistakes cancelled!
The expressions for the magnitudes are obtained from
##|\mathbf a + \mathbf b|^2=(\mathbf a + \mathbf b)\cdot(\mathbf a + \mathbf b)=a^2+b^2+2ab\cos\alpha##
##|\mathbf a - \mathbf b|^2=(\mathbf a - \mathbf b)\cdot(\mathbf a - \mathbf b)=a^2+b^2-2ab\cos\alpha##
where ##\alpha## is the angle between the two vectors when placed tail-to-tail, here 125°.

In the vector diagram, ##\vec S## is the difference, but the calculation below it is correct for the sum.
 
kuruman said:
In the vector diagram, ##\vec S## is the difference, but the calculation below it is correct for the sum.
The OP has incorrectly drawn the Post #1 diagram, believing (wrongly) that ##\vec s = \vec a + \vec b##.

Then they have ignored their diagram and used:
Ishfa said:
Relevant Equations: s= (a^2 + b^2 + 2ab cos alpha)^1/2
which gives the correct value.

The OP should note thast the question specifically says “By constructing vector diagrams, find the magnitudes and directions of …”.

If the OP had used their (incorrect) diagram correctly, they would have obtained ##|\vec S| \approx 8.4##.
 
Steve4Physics said:
The OP has incorrectly drawn the Post #1 diagram, believing (wrongly) that ##\vec s = \vec a + \vec b##.

Then they have ignored their diagram and used:

which gives the correct value.

The OP should note thast the question specifically says “By constructing vector diagrams, find the magnitudes and directions of …”.

If the OP had used their (incorrect) diagram correctly, they would have obtained ##|\vec S| \approx 8.4##.
That's another way of looking at it.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Back
Top