Two blocks of masses m and M are connected by means of metal wire of c...
Problem: Two blocks of masses m and M are connected by means of metal wire of cross section area A passing over a frictionless pulley the system is then released if M=2m then tension per unit crossectional area produce in wire?
Solution:Step 1: Free body diagram
First, draw the free body diagram of the system. The diagram will consist of two blocks, one pulley, and the wire connecting them. Assign the direction of motion for both the blocks.
Step 2: Applying Newton's Second Law
Apply Newton's Second Law of Motion to both the blocks. Since the pulley is frictionless, there is no force acting on it. Therefore, the tension in the wire will be the same on both sides of the pulley.
Step 3: Solving for Tension
Let T be the tension in the wire. Then, the equations of motion for the two blocks are:
mgsinθ - T = ma (1)
T - Mg = Ma (2)
where θ is the angle of inclination of the plane.
Substituting M=2m in equation (2), we get:
T - 2mg = 2ma
From equation (1), we get:
T = mgsinθ + ma
Substituting a from equation (2), we get:
T = mgsinθ + (T - 2mg)/2
Simplifying the equation, we get:
T = 2mgsinθ/3
Therefore, the tension per unit cross-sectional area produced in the wire is:
T/A = (2mgsinθ/3)/A
Step 4: Final Answer
Thus, the tension per unit cross-sectional area produced in the wire is (2mgsinθ/3)/A.