If a pushing force making an angle α with horizontal is applied on a block of mass m placed on horizontal table and angle of friction is β, then minimum magnitude of force required to move the block is

Question

If a pushing force making an angle α with horizontal is applied on a block of mass m placed on horizontal table and angle of friction is β, then minimum magnitude of force required to move the block is (A) (m g sin β/ cos [α-β]) (B) (m g sin β/ cos [α+β]) (C) (m g sin β/ sin [α+β]) (D) (m g cos β/ cos [α-β])

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/ab100bccb627d57cc56019feda96c177-.png" /> Angle of friction is β ⇒ μ= tan β N=m g +F sin α To just move the block F cos α=μ N F cos α= tan β(m g+ F sin α) F( cos α- tan β sin α)=m g tan β F( cos α cos β- sin α sin β)=m g sin β F=(m g sin β/ cos (α+β))

Q. If a pushing force making an angle $α$ with horizontal is applied on a block of mass $m$ placed on horizontal table and angle of friction is $β$ , then minimum magnitude of force required to move the block is

$\frac{m g s i n β}{c o s [ α - β ]}$

$\frac{m g s i n β}{c o s [ α + β ]}$

$\frac{m g s i n β}{s i n [ α + β ]}$

$\frac{m g c o s β}{c o s [ α - β ]}$

Angle of friction is $β$
$\Rightarrow μ = tan β$
$N = m g + F sin α$
To just move the block
$F cos α = μ N$
$F cos α = tan β (m g + F sin α)$
$F (cos α - tan β sin α) = m g tan β$
$F (cos α cos β - sin α sin β) = m g sin β$
$F = \frac{m g s i n β}{c o s ( α + β )}$

Q. If a pushing force making an angle α with horizontal is applied on a block of mass m placed on horizontal table and angle of friction is β, then minimum magnitude of force required to move the block is

cos[α−β]mgsinβ​

cos[α+β]mgsinβ​

sin[α+β]mgsinβ​

cos[α−β]mgcosβ​