Calculating Penguin Mass m2 with Forces and Frictionless Ice - Homework Help"

[dorkymichelle]

Homework Statement

The figure here shows four penguins that are being playfully pulled along very slippery (frictionless) ice by a curator. The masses of three penguins and the tension in two of the cords are m1 = 11 kg, m3 = 16 kg, m4 = 24 kg, T2 = 108 N, and T4 = 216 N. Find the penguin mass m2 that is not given.

http://edugen.wiley.com/edugen/courses/crs4957/art/qb/qu/c05/fig_5_D.gif

Homework Equations

F=ma

The Attempt at a Solution

total force =
t1+t2+t3+t4=(m1+m2+m3+m4)a
F=ma,
using penguin 4,
216N=24kg*a
a=9
t1+t2+t3+t4 =m1a+m2a+m3a+m4a
t1=(11)(9)=99
t3=(16)(9)=144
using
t1+t2+t3+t4=(m1+m2+m3+m4)a
99+108+144+216=99+m2(9)+144+216
108=m2(9)
m2 = 12
I realize I just did this the long way, instead of just doing F2=m2a but this is the wrong answer... how do I go about doing this

 

[ashishsinghal]

total force =
t1+t2+t3+t4=(m1+m2+m3+m4)a

This is wrong. Because:
On m1 force is t1
hence t1 = m1a
on m2
t2-t1=m2a
on m3
t3-t2=m3a
on m4
t4-t3 = m4a
On adding we get
t4 = (m1 + m2 + m3 + m4)a

 

[dorkymichelle]

So for each penguin after the first penguin, the force needed to pull those penguin includes the ones before it?

 

[ashishsinghal]

Yeah

 

[dorkymichelle]

ok using systems of equations
t2-t1=m2a
+ t1 = m1a
I got t2=a(m2a+m1a)
then I used
t3-t2=m3a
-t4-t3=m4a
= -t2-t4 = m3a-m4a
plugging numbers in, i got
-108-216=a(16-24+
-324=a(-8)
a= 40.5 m/s^2
then using the first equation i got
t2=a(m2+m1)
i got
108=(40.5)(m2)+445.5
and m=8.33 kg
but that's wrong too..

 

[ashishsinghal]

why have you done "-t4-t3=m4a" instead of "t4-t3=m4a"? Of course your answer will be wrong.

 

[dorkymichelle]

I did
t3-t2=m3a
- t4-t3=m4a
as a system of equations
so t3-(-t3)=0
and then -t2-t4 = m3a-m4a

 

[ileacus]

I did
t3-t2=m3a
- t4-t3=m4a
as a system of equations
so t3-(-t3)=0
and then -t2-t4 = m3a-m4a

Here's your problem:

T3 - (-T3) ≠ 0
T3 - (-T3) = T3 + T3

Try adding the above equations instead of subtracting.

BTW, you actually don't need to find a numerical value for the acceleration. You can rearrange the second derived equation (the one you get by adding the above equations) to get an expression for 'a', then substitute that into your first derived equation, T2 = a * (m1 + m2), and solve for m2.

 

[ashishsinghal]

I did
t3-t2=m3a
- t4-t3=m4a

Why have you done -t4 - t3=m4a. The direction of a is in the direction of t4 and opposite t3.
hence t4-t3 = m4a