Question

16. A block of mass m is kept on a plank
coefficient of friction between the plank and
is 1. The plank is slowly raised from one end
that it makes angle with horizontal. The forme
friction acting on the plank, when 0 = 30
0 = 60° are respectively,
Plank
mg
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was the man
(3) ma va mg
(2) Varme me
(a) me.me​

Answer 1

Answer:

plz mark me as brainliest

Answer 2

Answer:

A.$\frac{mg}{2},\frac{mg}{2}$

Explanation:

We are given that a  block of mass m is kept on a plank.

Coefficient of friction=1

We have to find the values of forces of friction acting on the plank when $theta=30^{\circ},\theta=60^{\circ}$

When the plank is raised slowly and makes an angle with horizontal=$\theta$

$tan\theta=\mu\implies tan\theta=tan45^{\circ}$

$\implies \theta=45^{\circ}$

b$tan45^{\circ}=1$

Angle of response=45 degrees

If angle is less than 45 degrees than the body does not move.

Then,  friction force when $\theta =30^{\circ}$

$f=mgsin30=mg\cdot \frac{1}{2}=\frac{mg}{2}$

$sin30^{\circ}=\frac{1}{2}$

When $\theta =60^{\circ}$

Friction force=$mgcos\theta$

Friction force=$mgcos60=\frac{mg}{2}$

$cos60^{\circ}=\frac{1}{2}$

Hence, option  A is true.

Answer:A.$\frac{mg}{2},\frac{mg}{2}$