Math Problem Statement

A block of mass M in slides on a frictionless surface. Find an expression for the tension in the string. Express your answer in terms of the variables m , M , and appropriate constants.

Solution

In this problem, we are likely dealing with two blocks connected by a string where one block is on a frictionless surface and the other is hanging over the edge, influenced by gravity. Here’s a common setup to derive the expression for the tension in the string:

Problem Setup:

  • A block of mass MM is on a frictionless horizontal surface.
  • Another block of mass mm is hanging vertically, connected to MM by a string passing over a frictionless pulley.
  • Both blocks are connected by the string, and as the block mm falls under gravity, it accelerates both blocks.

Step-by-Step Solution:

  1. Identify the forces acting on each block:

    • For the block of mass MM (on the horizontal surface):

      • It experiences no friction, so the only horizontal force is the tension TT in the string.
      • Newton's second law for MM: T = M a \tag{1} where aa is the acceleration of the system (the same for both blocks since they are connected by the string).
    • For the block of mass mm (hanging vertically):

      • It experiences two forces: the downward gravitational force mgmg and the upward tension TT in the string.
      • Newton's second law for mm: mg - T = m a \tag{2}
  2. Combine the equations: From equation (1), we have T=MaT = M a, and from equation (2), mgT=mamg - T = m a. Substituting T=MaT = M a into equation (2), we get: mgMa=mamg - M a = m a

  3. Solve for acceleration aa: mg=(M+m)amg = (M + m) a a = \frac{mg}{M + m} \tag{3}

  4. Find the tension TT: Now, substitute the value of aa from equation (3) into T=MaT = M a: T=M(mgM+m)T = M \left( \frac{mg}{M + m} \right) T=MmgM+mT = \frac{M m g}{M + m}

Final Answer:

The tension in the string is given by: T=MmgM+mT = \frac{M m g}{M + m}

This expression is in terms of the masses MM, mm, and the gravitational acceleration gg.


Would you like more details on this solution or have any questions?

Here are 5 related questions you might find useful to explore:

  1. How would the tension change if the mass mm is much larger than MM?
  2. What happens to the acceleration if MM becomes very large compared to mm?
  3. How would the solution change if there was friction on the surface?
  4. What would the tension be if both masses were equal?
  5. How does the tension change if the string is not massless?

Tip: When solving problems with connected objects, always treat the acceleration of the system as the same for all connected parts, unless stated otherwise.

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Math Problem Analysis

Mathematical Concepts

Newton's Second Law
Forces
Tension
Acceleration

Formulas

T = Ma
mg - T = ma
a = mg / (M + m)
T = Mmg / (M + m)

Theorems

Newton's Second Law

Suitable Grade Level

Grades 9-12