# Finding acceleration and tension

• tica86
In summary: Your summary should be something like this: "In summary, the problem involves calculating the acceleration, tension T1 and tension T2 in three different strings connected to three masses: one of 4 kg, one of 2 kg, and one of 1 kg. The coefficient of kinetic friction between each mass and the table is 0.1. The equations used to solve for these unknowns are based on the free body diagrams and Newton's laws. The equations are T1 -u(m1)g = (m1)a, T2 - T1 -u(m2)g = m2(a), and m3(g) - T2 = (m3)a. Solving these equations will give the values of acceleration
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!

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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)

tica86 said:
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

tica86 said:
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...

PhanthomJay said:
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?

tica86 said:
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?)

tiny-tim said:
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

tica86 said:
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.

PhanthomJay said:
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??

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.

PhanthomJay said:
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
??

tica86 said:
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.

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

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

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.

PhanthomJay said:
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.

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...

The legend said:
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.

what do you get adding all of the equations?

The legend said:
what do you get adding all of the equations?

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

it's wrong...

The legend said:
it's wrong...

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

cant help any further..

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The legend said:
cant help any further..

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

hi tica86!

(just got up :zzz: …)

as The legend says, there are two ways you can solve these three equations …
The legend said:
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 …
The legend said:
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
tiny-tim said:
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

tiny-tim said:
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 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..

The legend said:

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...

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..

## 1. What is acceleration?

Acceleration is the rate of change of an object's velocity over time. It is a vector quantity, meaning it has both magnitude and direction.

## 2. How do you calculate acceleration?

Acceleration can be calculated by dividing the change in velocity by the change in time. The formula for acceleration is a = (v2 - v1) / (t2 - t1), where v2 and v1 are the final and initial velocities, and t2 and t1 are the final and initial times.

## 3. What is tension?

Tension is a force that is transmitted through a string, rope, cable, or other similar objects when it is pulled tight by forces acting from opposite ends. It is measured in Newtons (N).

## 4. How do you find the tension in a system?

To find the tension in a system, you can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. By applying this law to the system and solving for tension, you can find the tension in the system.

## 5. Can tension affect acceleration?

Yes, tension can affect acceleration. If an object is being pulled by a string or rope, the tension in the string will contribute to the net force acting on the object, which will in turn affect its acceleration. The direction of the tension force will also affect the object's acceleration, as it is a vector quantity.

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