Vector 2B - Vector A (graphical method)

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Vector A is 4.00 units at 40 degrees below the x-axis, while Vector B is 3.00 units at 25 degrees below the negative x-axis. The task is to calculate Vector 2B - Vector A using a graphical method. Confusion arises regarding the components of the vectors and how to properly apply vector subtraction. After several attempts, one participant calculated the result as 2B - A = -0.8i - 0.6j, with a magnitude of 1, indicating ongoing uncertainty about the correct approach.
s31t8n8
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Homework Statement



Vector A is 4.00 units and 40* below x-axis. Vector B is 3.00 units and 25* below -x-axis. Use the graphical method (only ruler and protractor) to find Vector 2B - Vector A. I have to find the magnitude and direction.

When I did it I got 8.5 and East, but I'm not sure if I did it right. When they say 4.00 and 3.00 units, is that the ax component of the vector? That's the way I did it. I drew out the vector then measured 3 and 4 units on the ax comp. of both vectors, then measured from that point down to the vector and got the ay component. Then I just solved for the magnitude.

thanks
 
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s31t8n8 said:

Homework Statement



Vector A is 4.00 units and 40* below x-axis. Vector B is 3.00 units and 25* below -x-axis. Use the graphical method (only ruler and protractor) to find Vector 2B - Vector A. I have to find the magnitude and direction.

When I did it I got 8.5 and East, but I'm not sure if I did it right. When they say 4.00 and 3.00 units, is that the ax component of the vector? That's the way I did it. I drew out the vector then measured 3 and 4 units on the ax comp. of both vectors, then measured from that point down to the vector and got the ay component. Then I just solved for the magnitude.

thanks

The vector length is 4 at an angle of 40 degrees down below positive X axis and the other vector is 3 long pointing down from the negative x-axis. (The other direction of x)

But they say 2*B which makes that vector 6 long.
 
LowlyPion said:
The vector length is 4 at an angle of 40 degrees down below positive X axis and the other vector is 3 long pointing down from the negative x-axis. (The other direction of x)

But they say 2*B which makes that vector 6 long.

ok i just redrew it and now I am lost. I am getting confused about what theyre asking
 
s31t8n8 said:
ok i just redrew it and now I am lost. I am getting confused about what theyre asking

You've drawn the two vectors - doubling B?

Then reverse Vector A - same length - opposite direction.

They say 2*B - A right. This is vector subtraction. But you do that by "adding" the negative of the Vector A.

Now add these two vectors together.
 
ok i did it again and I am sure i got it wrong again but i got
2*B-A= -.8i-.6j
magnitude = 1
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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