An arrangement of three blocks is shown in the figure.The string and pulley are ideal.There is no friction between 6kg block and ground. Fricition between 5kg block and 6kg block is 0.8.The maximum value of mass m such that the two blocks of 6kg and 5kg will move together is?
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46
See diagram. As the diagram is not given, I have drawn the diagram, from the text of the problem. I hope that is the right one.
Equations of dynamics are:
m g - T = m a => T = m (g - a) ... (1)
T - μ * 5 * g = 6 a => T = 4 g + 6 a ... (2)
static friction = 5 a <= μ * 5 * g
=> a <= 0.8 g .... (3)
Solving (1) and (2):
m = (6a + 4g) / (g - a)
<= ( 6 *0.8 g + 4 g) / ( g - 0.8g)
<= 8.8/0.2
<= 44 kg
If m > 44 kg, then the block of 5 kg will slide backwards, as the maximum static friction between the two blocks, cannot accelerate it with the same acceleration of the 6 kg block.
Equations of dynamics are:
m g - T = m a => T = m (g - a) ... (1)
T - μ * 5 * g = 6 a => T = 4 g + 6 a ... (2)
static friction = 5 a <= μ * 5 * g
=> a <= 0.8 g .... (3)
Solving (1) and (2):
m = (6a + 4g) / (g - a)
<= ( 6 *0.8 g + 4 g) / ( g - 0.8g)
<= 8.8/0.2
<= 44 kg
If m > 44 kg, then the block of 5 kg will slide backwards, as the maximum static friction between the two blocks, cannot accelerate it with the same acceleration of the 6 kg block.
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Answered by
49
Answer:
The answer to this question is 22kg
Devide the whole system into 2
To make it easier
Explanation:
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