Question

The minimum value of coefficient of friction such that block of mass 5kg remains at rest is? please explain the answer briefly​

Answer 1

Answer:

Let the system be at rest.

For 3kg block

T - 3g = 0

so T = 3g. ............ (1)

For 5 kg block

1) For horizontal direction

T - f = 0

T = uN. ........... (2)

2) For vertical direction

5g = N. ............ (3)

From (1) (2) (3)

u(5g) = 3g

u = 3/5 = 0.6

So minimum coefficient of friction should be 0.6 so that 5 kg block remains at rest.

Answer 2

Given: Two blocks of mass 5kg and 3kg attached with a string

To find: The minimum value of the coefficient of friction such that block of mass 5kg remains at rest

Solution: Let the tension acting on string be T

for 5kg to remain at rest forces needs to be balanced on both blocks.

Normal force acting on 5kg will be equal to its weight that is 5g

equating forces in the horizontal direction on mass 5kg

T = friction acting on it

T = uN = u×5g

here u is the coefficient of friction

equating forces in the vertical direction on mass 3kg

T = 3g

So, we can say that 3g = u ×5g

u = 3/5

Therefore, the minimum value of the coefficient of friction such that block of mass 5kg remains at rest is 0.6.