A wooden wedge of mass M and inclination angle α rests on a smooth floor. A block of mass m is kept on wedge. A force overset arrow P is applied on the wedge as shown in figure, such that a block remains stationary with respect to wedge. The magnitude of force overset arrow P is

Question

A wooden wedge of mass M and inclination angle α rests on a smooth floor. A block of mass m is kept on wedge. A force overset arrow P is applied on the wedge as shown in figure, such that a block remains stationary with respect to wedge. The magnitude of force overset arrow P is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-kkbrspbiz5xhiq8t.jpg" /> (A) (M + m)g tan α  (B) g tan α  (C) mg cos α  (D) (M+m)g textcosec α

Tardigrade Admin · Accepted Answer

Since, P=(.M+m.)a Now as in free body diagram of block, <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-kkbrspbiz5xhv24c.jpg" /> ma cos α = m g sin â¡ α ∴ a=g( sin α / cos â¡ α )=g tan â¡ α or P=(M + m)g tan α

Q. A wooden wedge of mass $M$ and inclination angle $α$ rests on a smooth floor. A block of mass $m$ is kept on wedge. A force $P \to$ is applied on the wedge as shown in figure, such that a block remains stationary with respect to wedge. The magnitude of force $P \to$ is

$(M + m) g tan α$

$g tan α$

$m g cos α$

$(M + m) g cosec α$

Since, $P = (M + m) a$
Now as in free body diagram of block,

$ma cos α = m g sin α$
$∴ a = g \frac{s i n α}{c o s α} = g tan α$ or $P = (M + m) g tan α$

Q. A wooden wedge of mass M and inclination angle α rests on a smooth floor. A block of mass m is kept on wedge. A force P→ is applied on the wedge as shown in figure, such that a block remains stationary with respect to wedge. The magnitude of force P→ is

(M+m)gtanα

gtanα

mgcosα

(M+m)gcosecα