Two blocks ' A ' and ' B ' each of mass ' m ' are placed on a smooth horizontal surface. Two horizontal forces F and 2F are applied on the blocks A and B , respectively, as shown in figure. The block A does not slide on block B . The normal reaction acting between the two blocks is

Question

Two blocks ' A ' and ' B ' each of mass ' m ' are placed on a smooth horizontal surface. Two horizontal forces F and 2F are applied on the blocks A and B , respectively, as shown in figure. The block A does not slide on block B . The normal reaction acting between the two blocks is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-2hvcgtjp5dcw.png" /> (A) F (B) (F/2) (C) (F/√3) (D) 3F

Tardigrade Admin · Accepted Answer

Acceleration of two mass system is a=(F/2 m) leftward FBD of block A <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/55d95a66b3550f71e20fd4112c5a0e21-.png" /> Ncos 60o-F=ma ⇒ Ncos 60o-F=(m F/2 m) ⇒ N=3F

$F$

$\frac{F}{2}$

$\frac{F}{3}$

$3 F$

Acceleration of two mass system is $a = \frac{F}{2 m}$ leftward
FBD of block $A$

$N cos 6 0^{o} - F = ma$
$\Rightarrow N cos 6 0^{o} - F = \frac{m F}{2 m}$
$\Rightarrow N = 3 F$

Q. Two blocks ' A ' and ' B ' each of mass ' m ' are placed on a smooth horizontal surface. Two horizontal forces F and 2F are applied on the blocks A and B , respectively, as shown in figure. The block A does not slide on block B . The normal reaction acting between the two blocks is

F

2F​

3​F​

3F

Acceleration of two mass system is a=2mF​ leftward FBD of block A Ncos60o−F=ma ⇒Ncos60o−F=2mmF​ ⇒N=3F

$F$

$\frac{F}{2}$

$\frac{F}{3}$

$3 F$

Acceleration of two mass system is $a = \frac{F}{2 m}$ leftward
FBD of block $A$

$N cos 6 0^{o} - F = ma$
$\Rightarrow N cos 6 0^{o} - F = \frac{m F}{2 m}$
$\Rightarrow N = 3 F$