Solve Simple Force Question: Acceleration, Tension

  • Thread starter Thread starter Sheneron
  • Start date Start date
  • Tags Tags
    Force
Click For Summary
SUMMARY

The discussion addresses a physics problem involving a 2.00 kg object on a frictionless table connected to a 7.00 kg hanging object via a pulley. The solution involves setting up two equations based on Newton's second law, leading to the formula for acceleration: a = m2g / (m1 + m2). The confusion arises from the negative sign in the calculated acceleration, which is clarified by the definition of positive and negative directions in the problem. The correct interpretation is that a negative acceleration indicates the direction of motion relative to the defined coordinate system.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Basic knowledge of forces and tension in strings
  • Familiarity with coordinate systems in physics
  • Concept of gravitational acceleration (g = 9.8 m/s²)
NEXT STEPS
  • Review Newton's second law and its applications in dynamics
  • Study the concepts of tension in strings and pulleys
  • Learn about coordinate systems and their impact on motion analysis
  • Explore examples of free fall and acceleration in different contexts
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to forces and motion.

Sheneron
Messages
360
Reaction score
0
[SOLVED] Simpe Force Question

Homework Statement


A 2.00 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 7.00 kg object, as in Figure P5.24. Find the magnitude of the acceleration of the two objects and the tension in the string.

The Attempt at a Solution


I have a specific question about this problem that I came upon at the end. Here is how I solved it.

[tex]T = m_1a[/tex]
[tex]-T + m_2g = m_2a[/tex]
So two equations to uknowns, and here is what I got.
[tex]-m_1a + m_2g = m_2a[/tex]
[tex]a = \frac{m_2g}{(m_1+m_2)}[/tex]

So solving for a I got -7.62. The thing I have a question about is the negative on the acceleration. At the beginning of the problem I defined rightward and downward direction as positive and upward direction as negative. So if acceleration is a negative value does that not mean it is traveling up, since I defined up as negative? I know that it doesn't not, but I am confused it can be this way.
 
Physics news on Phys.org
Hi Sheneron,

How did you get a negative answer? All the values (masses and g) are positive numbers.

Did you use g=-9.8? The value of g is a magnitude and therefore always a positive number; if we were talking about an object in free fall we would say that its acceleration is

[tex] a_y = -g[/tex]

just to keep g a positive number.
 
Ah ok I see. thanks
 

Similar threads

Replies
44
Views
4K
Replies
25
Views
4K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 18 ·
Replies
18
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
13
Views
2K