Two blocks of masses is m and 2m are held in the equilibrium on a frictionless incline. In terms of m and x, find the magnitude of the tension t1 in the upper cord and the magnitude of the tension t2 in the lower cord connecting two blocks.
To find the magnitude of the tension in the upper cord (t1) and the lower cord (t2), we can analyze the forces acting on the blocks.
Let's break down the problem step by step:
1. Draw a free-body diagram for each block:
Block of mass m:
- The force of gravity (mg) acts vertically downward.
- The tension in the upper cord (t1) acts vertically upward.
- The normal force (N) acts perpendicular to the incline.
- There is no force of friction since the incline is assumed to be frictionless.
Block of mass 2m:
- The force of gravity (2mg) acts vertically downward.
- The tension in the lower cord (t2) acts vertically upward.
- The normal force (2N) acts perpendicular to the incline.
- There is no force of friction since the incline is assumed to be frictionless.
2. Resolve the forces along the incline:
Since the incline is frictionless, the only force component along the incline is the component of the force of gravity acting parallel to it.
For the block of mass m:
- The component of the force of gravity acting along the incline is m * g * sin(theta), where theta is the angle of incline.
For the block of mass 2m:
- The component of the force of gravity acting along the incline is 2m * g * sin(theta).
3. Apply Newton's second law (F = ma):
For the block of mass m:
- In the vertical direction, summing the forces gives: t1 - mg = 0
Therefore, t1 = mg.
For the block of mass 2m:
- In the vertical direction, summing the forces gives: t2 - 2mg = 0
Therefore, t2 = 2mg.
So, the magnitude of the tension in the upper cord (t1) is mg where m is the mass of the block, and the magnitude of the tension in the lower cord (t2) is 2mg where m is the mass of the block.