How Did This Happen | Math Forums

Math Forums

• Home High School Math Elementary Math Algebra Geometry Trigonometry Probability & Statistics Precalculus University Math Calculus Linear Algebra Abstract Algebra Applied Math Advanced Probability & Statistics Number Theory Topology Real Analysis Complex Analysis Differential Equations Math Discussions Computer Science Economics Math Software Math Books STEM Guidance
• Forums
• Trending
• Engineering
• Physics
• Chemistry

Login Register

Search

Everywhere Threads This forum This thread Search titles only By: Search Advanced search… Everywhere Threads This forum This thread Search titles only By: Search Advanced… Menu Install the app Install How to install the app on iOS

Follow along with the video below to see how to install our site as a web app on your home screen.

Note: this_feature_currently_requires_accessing_site_using_safari

• Home
• Physics
• Physics

You are using an out of date browser. It may not display this or other websites correctly.You should upgrade or use an alternative browser. How did this happen

• Thread starter r-soy
• Start date Dec 9, 2010
• Tags happen

R

r-soy

Joined Oct 2009 895 Posts | 1+ Hi all I want your help equation (1) a = g(m1+m2)/m1+m2 equation (2) T =m2g + m2g From equation (1) + 2 T = 2m1m2g/m1+m2 this equation 3 How do we get the equation 3

MarkFL

Joined Jul 2010 12K Posts | 548+ St. Augustine, FL., U.S.A.'s oldest city You have copied the equations incorrectly. I am assuming, from the other problem you posted, that you are analyzing Atwood's machine. When two unequal masses are hung vertically over a frictionless pulley of negligible mass, the arrangement is called Atwood's machine. Two forces act upon each mass: the upward force exerted by the string, the tension T, and the downward force of gravity. Thus, the magnitude of the net force exerted on m_1 is T-m_1g , while the magnitude of the net force exerted on m_2 is T-m_2g . Because the masses are connected by a string, their accelerations must be equal in magnitude. If we assume that m_2>m_1 , then m_1 must accelerate upward, while m_2 must accelerate downward. When Newton's second law is applied to m_1 , with upward acceleration a, we find (taking upward to be the positive y-direction) (1) \sum F_y=T-m_1g=m_1a Similarly, for m_2 we find (2) \sum F_y=T-m_2g=-m_2a The negative sign on the right-hand side of (2) indicates that m_2 accelerates downward, in the negative y-direction. When (2) is subtracted from (1), T drops out and we get: -m_1g+m_2g=m_1a+m_2a or (3) a=\left(\frac{m_2-m_1}{m_1+m_2}\right)g When (3) is substituted into (1), we get (4) T=\left(\frac{2m_1m_2}{m_1+m_2}\right)g The result for the acceleration, (3), can be interpreted as the ratio of the unbalanced force on the system to the total mass of the system. Login or Register / Reply

Similar Discussions

W Simple Harmonic Motion

• Waitzkin
• Nov 21, 2025
• Physics

Replies 10 Views 210 Physics Nov 30, 2025 Waitzkin W W The speed of EM waves is the speed of light

• Waitzkin
• Nov 11, 2025
• Physics

Replies 6 Views 175 Physics Nov 13, 2025 Waitzkin W W My First Solution to a Differential Equation

• Waitzkin
• Nov 24, 2025
• Differential Equations

Replies 4 Views 135 Differential Equations Nov 25, 2025 SDK S R The Zero Property of the Lorentz Transformation

• Reukauf
• May 2, 2025
• Physics

Replies 1 Views 582 Physics Jun 5, 2025 Matt C Anderson M

• Home
• Physics
• Physics