Two masses on two inclined planes

[maysenator]

Homework Statement

If the left hand slope makes a 60 degree angle and the right hand slope makes a 20 degree angle.
the left hand mass is 2.1 kg, and the right hand mass is unknown.

what would the right hand mass be given a downward acceleration of .64 m/s^2. ?

upslope at .76 m/s^2

Homework Equations

I have set this problem up to the best of my knowledge and watched tons of videos but cannot grasp it. i think I am not breaking it up enough and i have a picture of my work and how i set it up.

The Attempt at a Solution

thank you in advance for any help anyone may have to offer

thank you

 

[tiny-tim]

welcome to pf!

hi maysenator! welcome to pf!

(try using the X² icon just above the Reply box )

call the tension T, and the unknown mass m,

and apply good ol' https://www.physicsforums.com/library.php?do=view_item&itemid=26" (F = ma) twice, once for each mass, in the direction of the slope …

that gives you two equations with two unknowns, so you can solve for m (and T) …

what do you get? ​

 

[maysenator]

im a little confused, i don't see how you would do this without knowing the force on the side with the unknown mass,

just get 1.34 N for the known side?

 

[tiny-tim]

hi maysenator!

im a little confused, i don't see how you would do this without knowing the force on the side with the unknown mass,

call the unknown mass m …

you'll get a force with m in it …

what is it? ​

 

[maysenator]

2.09?

 

[maysenator]

my answer sheet says that mass of the second block should be 7.07 kg

 

[maysenator]

my ADHD is never going to let me take in all that i need to to pass this physics class. i can't keep up with so many things going on. in my head it feels like there is something i cannot see that others can. i have been sitting here since 10am and its 3pm now on this one problem about to cry. it shouldn't be this hard i just wish someone could see through my eyes because i don't know what to do

 

[tiny-tim]

you'll get a force with m in it …

what is it? ​

2.09?

2.09 what?

and why doesn't it have an m in it?

(and how did you get 2.09 anyway?)

 

[maysenator]

1.34/.64 = 2.09

f = ma?

 

[tiny-tim]

1.34/.64 = 2.09

f = ma?

that's 1.34 from the left hand mass?

no, your F = ma equation must be for one mass only …

do this slowly and logically …

first thing is to draw a https://www.physicsforums.com/library.php?do=view_item&itemid=100" with all the forces on it

(it must always be only for the forces on that body)

there are three forces on the right hand mass:

the weight 9.8m

the tension T

the normal force​

you know the acceleration parallel to the slope is 0.64

so now look at the diagram, work out what the component of the weight along the slope is, and write F = ma for that direction ​

 

[maysenator]

are you saying it SHOULD be for one mass only or that was my mistake?? sorry

again, thank you so much i feel closer already.

this is what i have drawn out so far

 

[tiny-tim]

no, your free body diagram for the right hand mass should be a slope at 20° with a little box on it

then three arrows from the box, one vertically down marked 9.8m, one along the slope marked T, and one perpendicular to the slope marked N

finally mark the acceleration of 0.64 down the slope (but don;t attach it to the box, and draw it differently from the forces, since it's not a force, so it shouldn't really be there! )

(and yes, it SHOULD be for one mass only)

 

[maysenator]

but but but i can't find the tension with this information>? i need to use information from the other diagram right??

 

[maysenator]

I don't get what I am missing i have done everything you have said ( i think ) but on my paper i don't have enough information to do anything with?

 

[tiny-tim]

but but but i can't find the tension with this information>? i need to use information from the other diagram right??

yes, you need to do a free body diagram, and F = ma, twice, once for each mass

(i thought you'd already done it for the left hand mass)

(the left hand one should give you T, then you put that value of T into the right hand one, to find m)

what are your two F = ma equations?​

 

[maysenator]

i didnt use f = ma for the left one. I thought that 20.58N * sin60 would give the force parallel to the slope

getting 17.83 and then 20.58 cos 60 = 10 . 29 N would the the normal force

so what... then i have ... Fnet = ma~

t1 + 20.58 + 10.29 = (2.1)(.64)
(all the forces?)

is that right? >.<? ( for the left )

 

[tiny-tim]

hi maysenator!

so what... then i have ... Fnet = ma~

t1 + 20.58 + 10.29 = (2.1)(.64)
(all the forces?)

ah, no …

you're fundamentally misunderstanding the vector nature of F = ma …

force and acceleration are both vectors, and F = ma is a vector equation …

that means you must choose a direction, and take components of F and of a in that direction

try it again, doing it in the direction of the slope ​

(btw, no need to call it T₁, then two tensions are the same, so just call it T )

 

[maysenator]

i don't understand what you mean.

the block is moving up the slope at .64 ms

its Fg is 9.8 * 2.1 right?

so i have a, and fg right there.

the normal force is the force the slope is pushing back up into the mass right?

moving in the x direction is Fg * cos 60 right? or Fg sin 60??

so why not plug it in from there?

T + fg + Fn = (2.1)(.64)

 

[maysenator]

thanks for your help man, I am giving up.. dropping the class going to try again next semester

paace

 

[maysenator]

great i can't drop this class because then ill lose my financial aid.

time to commit suicide

 

[tiny-tim]

hi maysenator!

(just got up :zzz: …)

i don't understand what you mean.

the block is moving up the slope at .64 ms

its Fg is 9.8 * 2.1 right?

so i have a, and fg right there.

the normal force is the force the slope is pushing back up into the mass right?

moving in the x direction is Fg * cos 60 right? or Fg sin 60??

so why not plug it in from there?

T + fg + Fn = (2.1)(.64)

because you must use the components (along the slope), not the whole forces

there are three forces on the left hand mass:

the weight W, the normal force N, and the tension T

ok let's do this slowly and logically …

i] what is the component of W along the slope (up being +ve)?

ii] what is the component of N along the slope (up being +ve)?

iii] what is the component of T along the slope (up being +ve)? ​

 

[maysenator]

just wanted to pop back in here and say thanks for the help, was just getting frustrated. ended up getting a decent grade in the class thanks to your help and others from this site.

never taking a calc based phys course during the summer again. I need more time to learn LOL