Question

q 13 plz solve with proof​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Same Force P is applied.

$\huge{\bold{\underline{Explanation:-}}}$

Let the coffecient of friction be ${ \mu}$

Case-1

So, Frictional Force will come as:-

$\large{N = Mg}$

By the friction formula.

$\large{\bold{f = \mu N}}$

Substituting the values.

$\large\boxed{\boxed{ \implies f = \mu Mg}}$

Case-2

So, Frictional Force will come as:-

$\large{N = Mg + P \sin \theta}$

By the friction formula.

$\large\bold{f = \mu N}$

Substituting the values.

$\large\boxed{\boxed{ \implies f = \mu (Mg + P \sin \theta)}}$

Case-3.

So, Frictional Force will come as:-

$\large{N = Mg - P \sin \theta}$

By the friction formula.

$\large{\bold{f = \mu N}}$

Substituting the values.

$\large\boxed{\boxed{ \implies f = \mu (Mg - P \sin \theta)}}$

Case-4.

As force is acting upwards

Frictional force will be.

$\large\boxed{\boxed{f = P (external \: Force \: applied)}}$

In above cases we saw that in case - 3 the frictional force is minimum compared to all the cases given in the question.

So, It will be easier to move the body in case-3.

#For figure of cases 2 and 3 refer the attachment.