Two blocks of masses M = 10 kg and m = 5kg are kept on a smooth horizo...
Given information:
- Mass of block M: M = 10 kg
- Mass of block m: m = 5 kg
- Pull force: F = 30 N
- Angle of pull force with respect to horizontal: θ = 60°
Assumptions:
- The surface is smooth, which means there is no friction between the blocks and the surface.
- The string connecting the two blocks is massless and inextensible.
Analysis:
To find the tension in the string connecting the two blocks, we need to consider the forces acting on each block separately and then apply Newton's second law of motion.
For block M:
- The only force acting on block M is the pull force F, which is at an angle of 60° with respect to the horizontal.
- Resolving the force F into its horizontal and vertical components, we get:
- Fx = F * cos θ = 30 * cos 60° = 30 * 0.5 = 15 N (horizontal component)
- Fy = F * sin θ = 30 * sin 60° = 30 * √3/2 = 15√3 N (vertical component)
For block m:
- The tension in the string connecting the two blocks acts on block m.
- Since the string is massless and inextensible, the tension is the same throughout the string.
- The weight of block m acts vertically downwards and can be calculated as:
- Weight = mass * acceleration due to gravity = m * g = 5 * 9.8 = 49 N
Applying Newton's second law:
- For block M:
- Sum of forces in the horizontal direction = Fx = mass * acceleration
- Since the surface is smooth, there is no acceleration in the horizontal direction.
- Therefore, Fx = 0, which means the horizontal component of the pull force is balanced by the tension in the string.
- Tension = Fx = 15 N
- For block m:
- Sum of forces in the vertical direction = Fy - weight = mass * acceleration
- Since the block is in equilibrium in the vertical direction, the sum of forces is zero.
- Therefore, Fy - weight = 0
- Tension = Fy = weight = 49 N
Conclusion:
The tension in the string connecting the two blocks is 15 N.
Two blocks of masses M = 10 kg and m = 5kg are kept on a smooth horizo...
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