Finding acceleration and tension

[tica86]

I have a test coming up and I want to be able to understand the following problem:

I attached an image.

The large crate has a mass of 4kg and the smaller crate has a mass of 2kg. The coefficient of kinetic friction between each crate and the wooden table is 0.1. A hanging weight of mass 1 kg causes both crates to accelerate to the right. What's the acceleration of the block and the tension T1 and T2 in strings A and B??

My attempt:
I think I would have to set up three different equations to find acceleration?

Mass 1:
T-0.1=4a

Mass 2:
2mg+T1+T2=2mg

Mass 3:
T-1g=-1a

Can anyone help me out, please. Thanks!

 

[tiny-tim]

hi tica86!

if all you wanted was the acceleration, you could just treat the whole thing as one-dimensional (along the string), and have one equation with no tension

but since the question specifically asks for the tensions, yes, you need three equations (either three F = ma equations, or two of them plus the one-dimensional equation)

Mass 1:
T-0.1=4a

Mass 2:
2mg+T1+T2=2mg

Mass 3:
T-1g=-1a

equation 3 is fine

equation 1 you need to use µN not just µ !

but sorry i don't understand what equation 2 is

 

[PhanthomJay]

I have a test coming up and I want to be able to understand the following problem:

I attached an image.

The large crate has a mass of 4kg and the smaller crate has a mass of 2kg. The coefficient of kinetic friction between each crate and the wooden table is 0.1. A hanging weight of mass 1 kg causes both crates to accelerate to the right. What's the acceleration of the block and the tension T1 and T2 in strings A and B??

My attempt:
I think I would have to set up three different equations to find acceleration?

yes, correct.

Mass 1:
T-0.1=4a

you forgot to myultiply the coef of friction by m1(g)

Mass 2:
2mg+T1+T2=2mg

T2 acts right, T1 and the friction force act left, and the acceleration is 'a', not '2g'.

Mass 3:
T-1g=-1a

That T is T3, whch is equal to T2 in magnitude. You should probably then rewrite it as 1g - T2 = 1a.

Edit: Tiny Tim was a wee bit faster, but we concur...

 

[tica86]

yes, correct. you forgot to myultiply the coef of friction by m1(g) T2 acts right, T1 and the friction force act left, and the acceleration is 'a', not '2g'. That T is T3, whch is equal to T2 in magnitude. You should probably then rewrite it as 1g - T2 = 1a.

Edit: Tiny Tim was a wee bit faster, but we concur...

Ok I just want to make sure:

Mass 1:
T-0.1(+9.8)=4a?
so T-.98=4a

Mass 2:
I'm sorry I don't understand how I would set up the equation for mass 2
would it be
T2+9.8=2a?

Mass 3:
1g - T2 = 1a?

 

[tiny-tim]

Mass 2:
I'm sorry I don't understand how I would set up the equation for mass 2
would it be
T2+9.8=2a?

i'm still not sure what you're doing here

if this is an F = ma equation, then you need all three of the horizontal forces on mass 2 (and where's your µ?)

 

[tica86]

i'm still not sure what you're doing here

if this is an F = ma equation, then you need all three of the horizontal forces on mass 2 (and where's your µ?)

Ok, so I would need friction and T1?

-T1+T2-.98=2a??

I'm really confused with trying to set up this equation.
I put -T1 because it's acting to the left and +T2 because it's
acting to the right.
-.98 because friction moves in the opposite direction of motion
so -1.0(9.8)=-.98

 

[PhanthomJay]

Ok, so I would need friction and T1?

-T1+T2-.98=2a??

I'm really confused with trying to set up this equation.
I put -T1 because it's acting to the left and +T2 because it's
acting to the right.
-.98 because friction moves in the opposite direction of motion
so -1.0(9.8)=-.98

But the block m2 has a mass of 2 kg, , and u =0.1, so the friction force is___? Otherwise, you are on track. Check the friction force on mass 1 also.

 

[tica86]

But the block m2 has a mass of 2 kg, , and u =0.1, so the friction force is___? Otherwise, you are on track. Check the friction force on mass 1 also.

so mass 2:

-T1+T2-0.2=2a?
I really don't think that's right :/

Mass 1:
T-0.1(4)=4a?
T-0.4=4a??

 

[PhanthomJay]

Ok let's get right to the point: Kinetic friction force = (u_k)(N), where N is the normal force (N), where N is equal to the crate's weight (mg) in this example.

 

[tica86]

Ok let's get right to the point: Kinetic friction force = (u_k)(N), where N is the normal force (N), where N is equal to the crate's weight (mg) in this example.

ok, so
m1=0.1(39.2)=3.92
m2=0.1(19.6)=1.96
??

 

[PhanthomJay]

ok, so
m1=0.1(39.2)=3.92
m2=0.1(19.6)=1.96
??

Friction on mass m1 is 3.92 N acting left.
Friction on m2 is 1.96 N acting left.

 

[tica86]

Friction on mass m1 is 3.92 N acting left.
Friction on m2 is 1.96 N acting left.

So both answers are negative?
m3=1(9.8)? I wouldn't add friction here correct?
so after that to find acceleration
a=Fnet/m
so Fnet=m3g-5.88?? This is how I got -5.88: -3.92 + -1.96
m3g=-9.8?
-----------m=39.2(m1)+19.6(m2)+9.8(m3)=68.6

 

[PhanthomJay]

You are confusing mass and weight, and a lot of other things.

You started off on the right foot when you tried to develop three equations , one associated with each mass. These equations are developed using free body diagrams and Newton's laws. Then you went wild.

Looking at the first mass, T1 acts right and friction acts left. So you have F_net =T1 -u(m1)g = (m1)a.

looking at the 2nd mass, you have T2 - T1 -u(m2)g = m2(a)

looking at the third mass, you have m3(g) -T3 = (m3)a, but since tensions on both sides of an ideal pulley are the same in magnitude, then T3 =T2, and your 3rd equation becomes

m3(g) - T2 = (m3)a

Thus, you now have 3 equations with 3 unknows (T1, T2, and a). Solve for all unknowns using the solution of your choice. For simplicity, plug in the values you know before solving.

 

[tica86]

You are confusing mass and weight, and a lot of other things.

You started off on the right foot when you tried to develop three equations , one associated with each mass. These equations are developed using free body diagrams and Newton's laws. Then you went wild.

Looking at the first mass, T1 acts right and friction acts left. So you have F_net =T1 -u(m1)g = (m1)a.

looking at the 2nd mass, you have T2 - T1 -u(m2)g = m2(a)

looking at the third mass, you have m3(g) -T3 = (m3)a, but since tensions on both sides of an ideal pulley are the same in magnitude, then T3 =T2, and your 3rd equation becomes

m3(g) - T2 = (m3)a

Thus, you now have 3 equations with 3 unknows (T1, T2, and a). Solve for all unknowns using the solution of your choice. For simplicity, plug in the values you know before solving.

I wasn't kidding when I said I'm confused, I haven't taken a math course in years so solving might be a bit hard for me, let me try :/
Mass 1:
T1-3.92=4a

Mass 2:
T2-T1-1.96=2a

Mass 3:
9.8-T2=1a

To find acceleration would I add up all the same variables so:
T2-T1+7.84=7a?

I'm sorry I know it must be really annoying that I keep asking questions but if I don't I'm not going to understand let alone do well on my test.

 

[The legend]

not exactly...
first eliminate T1 from the 1st equation with help of 2nd
Then you have 2 variables that's T2 and a and then eliminate T2 from 2nd equation with help of third... hence get a ... substitute and solve then...

 

[tica86]

not exactly...
first eliminate T1 from the 1st equation with help of 2nd
Then you have 2 variables that's T2 and a and then eliminate T2 from 2nd equation with help of third... hence get a ... substitute and solve then...

I understand what you are saying but I don't know how I would do that, and my previous attempts have been wrong.

 

[The legend]

what do you get adding all of the equations?

 

[tica86]

what do you get adding all of the equations?

doubt this will be correct...
T2-T1-5.88=6a?

 

[The legend]

it's wrong...
cmon it's just addition (add all the three equations)

 

[tica86]

it's wrong...
cmon it's just addition (add all the three equations)

that's what I tried doing previously but was told it was wrong, if I'm adding all the
like variables, this is what I get:
T2-T1+3.92=7a

 

[The legend]

cant help any further..

 

[tica86]

cant help any further..

would have liked to be able to solve it, thanks anyways

 

[tiny-tim]

hi tica86!

(just got up :zzz: …)

as The legend says, there are two ways you can solve these three equations …

first eliminate T1 from the 1st equation with help of 2nd
Then you have 2 variables that's T2 and a and then eliminate T2 from 2nd equation with help of third... hence get a ... substitute and solve then...

surely you've done this before (eliminating one unknown at a time, so that you get down from having three unknowns in three equations to having one unknown in one equation)? if not, you do need to practise it, it'll come up a lot

alternatively, in this particular case, you can just …

what do you get adding all of the equations?

because if you do that, the +T₁ cancels out the -T₁, and the T₂ cancels out the -T₂ …

in fact, that gives you the equation i mentioned at the start …

if all you wanted was the acceleration, you could just treat the whole thing as one-dimensional (along the string), and have one equation with no tension

try both methods …

show us what you get ​

 

[tica86]

hi tica86!

(just got up :zzz: …)

as The legend says, there are two ways you can solve these three equations …


surely you've done this before (eliminating one unknown at a time, so that you get down from having three unknowns in three equations to having one unknown in one equation)? if not, you do need to practise it, it'll come up a lot

alternatively, in this particular case, you can just …


because if you do that, the +T₁ cancels out the -T₁, and the T₂ cancels out the -T₂ …

in fact, that gives you the equation i mentioned at the start …


try both methods …

show us what you get ​

Ok well it's been years since I've taken an algebra course so that's why I'm having a hard time trying to solve.
This are the answers that I got, I used a very long elimination method so if it is correct can you let me know a shorter way to solving it, thanks tiny-tim!

a=0.56
T2=9.24
T1=6.16

 

[The legend]

^ your answers look correct ... nice!

The way i told you to add up all of them is really, really easy and isn't very long...
by a little practice you can do such sums orally..

 

[tica86]

^ your answers look correct ... nice!

The way i told you to add up all of them is really, really easy and isn't very long...
by a little practice you can do such sums orally..

If you are able can you show me how you would do that, each step? if u can...

 

[The legend]

adding all
T1 cancels with the -T1 and T2 with -T2
so
9.8-3.92-1.96=7a
solving
a=0.56

now , substituting a in 1st and third equn you get the answer..

Practice, and you can do this orally, as i already said..