Question

Find the magnitude of the force needed to accelerate a 220g mass with a⃗ = -0.145m/s2 i^+0.430m/s2 j^.

Answer 1

you cant find the magnitude of the acceleration vectors by using the pathagorean : (-.145)^2+(.430)^2=a^2 and solve for a and multiply this value by the mass in kilograms.

Answer 2

can*

Answer 3

\[\sqrt{(-.145)^{2}+(0.43)^{2}}=a(220*10^{-3})=\left| F \right| \]

Answer 4

Can you just multiply the mass times the acceleration?

Answer 5

yes, but you need the net acceleration. Take a look at the equation above.

Answer 6

In that equation there are two variables, the a and the F?

Answer 7

is the acceleration the answer in parenthesis?

Answer 8

Sorry, I wrote the equation wrong. Once you solve for a, then you can multiply by the mass.

Answer 9

the accelleration is computed with the pathagorean, and then you multiply the term in parentheses (mass converted to kilograms) to find the force. I'll rewrite the equation properly.

Answer 10

I got .0999 N. is that correct?

Answer 11

\[\sqrt{a _{x}^{2}+a _{y}^{2}}=a _{net},\left| F \right| =ma _{net}\]

Answer 12

Not sure, let me calculate.

Answer 13

that's it.

Answer 14

and if i were to find the angle of this do i take the tangent of it?

Answer 15

sorry not angle but direction

Answer 16

you would take the arctangent of the two acceleration component vectors.

Answer 17

You can figure out direction just by looking at the signs of the component vectors for the acceleration vector in relation to the cartesian coordinate grid.

Answer 18

would the answer be 15.02 im not sure what the unit is

Answer 19

no, I got the value 18.6 degrees.

Answer 20

how do you convert that answer using dimensions of plane angle.

Answer 21

I don't know what that means. What is plane angle?

Answer 22

well the question is asking for the direction of the force
\[\theta=\]

Answer 23

then you would make the degree angle negative.

Answer 24

Hey, sorry gave you the wrong angle earlier. I got -71.4 degrees

Answer 25

\[\tan^{-1} (0.43/-0.145)=\theta \]

Answer 26

i tried it and got the same answer to but it keeps saying its wrong, im not sure what exactly its asking for

Answer 27

maybe it wants 108.6 degrees. since the vector should be in the second quadrant of the cartesian

Answer 28

Yes that is the correct answer. How did you know it was suppose to be in the second quadrant

Answer 29

just by looking at the vector signs. -+ is the second quadrant of the cartesian. -x, +y

Answer 30

Awesome. I get it now. Thanks. :D

Answer 31

no problem.