Question

The coefficient of friction between two surfaces is
u = 0.8. The tension in the string

(a) ON
(0) 4N
(b) 6N
(d) 8N​

Answer 1

Given:

The frictional force coefficient μ=0.8

To find:

The tension in the string

Solution:

The normal is acting upward and weight acting downward

As F=μN=μmgcosФ

F=0.8×1×10×cos(30)

F=6.928 N

So now the frictional also acting downward direction

mgsinФ=1×10×sin30

=5 N

So the downward force

mgsinФ=F+T

6.928=5+T

T=0

We have taken zero as the approx value of the tension on the string.

Option A is correct

Answer 2

Answer:

(a) 0N

Explanation:

Maximum frictional force acting on the block   f=μmg cos30o

f=0.8×1×9.8×0.866=6.8 N

Downward force  =mg sin30=1×9.8×0.5=4.9 N

Since downward force is less than maximum frictional force, thus the tension in the string is zero.