two blocks of the mass M kg and 2,5 kg respectively are connected by a light in extensible string. the string runs over a light frictionless pulley. the blocks are stationary.
the coefficient of static friction between the unknown mass M and the surface of the table is 0,2.
a)calculate the minimum value of M that will prevent the blocks from moving.
First, we need to consider the forces acting on the two blocks.
For the block with mass M:
- Tension in the string
- Weight force acting downwards
- Normal force acting upwards
- Friction force acting to the left
For the block with mass 2.5 kg:
- Tension in the string
- Weight force acting downwards
Since the blocks are stationary, the forces in the vertical direction must balance each other out:
For the block with mass M:
N - Mg = 0
N = Mg
For the block with mass 2.5 kg:
T - 2.5g = 0
T = 2.5g
Now, considering the horizontal direction for the block with mass M:
Ffriction = T - fstatic
fstatic = μN
fstatic = 0.2Mg
Since the blocks are on the verge of moving:
fstatic = T
0.2Mg = 2.5g
0.2M = 2.5
M = 2.5 / 0.2
M = 12.5 kg
Therefore, the minimum value of M that will prevent the blocks from moving is 12.5 kg.