Newton's law of motion problem

  • Thread starter kfink85
  • Start date
  • #1
12
0
Here is the problem. I think I set it up right but I am stumped now.

An object with mass m1=5.00 kg rests on a frictionless horizontal table and is connected to a cable that passes over a pulley and is then fastened to a hanging object with mass m2=10.0 kg. Find the acceleration of each object and the tension in the cable.

Here is what I came up with:

m1y: N-m1*g=0
N=m1*g
N=-49N

m1x: T=m1*ax
ax=T/m1

m2: T-m2*g=m2*a
(T-m2*g)/m2=a

Help please!!! This is due tomorrow and I'm stuck. I'm really bad at solving or finding the right step to get to only 1 unknown in equations.
 

Answers and Replies

  • #2
daniel_i_l
Gold Member
867
0
First of all - the N is positive = mg.
Now with each of the blocks you have an equation with two unknowns, the acceleration which is the same with both blocks and the tension which is also the same in both equations. Solving this is just simple algebra. You could do it by using the first equation to find T as a function of a (T=m1*a) and then inserting that (m1*a) into the second equation. Then you will have just one unknown.
 
  • #3
759
0
You know that the magnitude of acceleration is the same for both masses?

Therefore,
[tex] | {\vec a} | = \frac{{| {m_2 \vec g - \vec T} |}}{{m_2 }} = \frac{{ | {\vec T} | }}{{m_1 }} \Rightarrow[/tex]

[tex]| \vec T | = \frac{{m_1 m_2 }}{{m_1 + m_2 }}| \vec g |[/tex]

[tex]| \vec a | = \frac{{m_2 }}{{m_1 + m_2 }} | \vec g |[/tex]

Now just plug in that m1=5.00 kg and m2=10.0 kg //
(and that [itex]| \vec g |[/itex] = 9.81 m/s2 :smile:)
 
Last edited:
  • #4
12
0
Thanks very much guys.
 

Related Threads on Newton's law of motion problem

  • Last Post
Replies
4
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
3
Views
2K
Replies
5
Views
7K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
1
Views
6K
Replies
8
Views
4K
Replies
2
Views
4K
Replies
2
Views
6K
Top