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4-1.Newton's Laws of Motion
normal
A perfect smooth sphere $A$ of mass $2\ kg$ is in contact with a rectangular block $B$ of mass $4\ kg$ and vertical wall as shown in the figure. All surfaces are smooth. Find normal reaction by vertical wall on sphere $A$ .......... $N$
A$20$
B$25$
C$80$
D$45$
Solution
Balancing forces
$M_{A} g \cos 37+N_{\text {wall }} \cos 53=N_{A}$
$N_{A B}+M_{A} g \sin 37=N_{\text {wall }} \sin 53$
$N_{A B}=M g \sin 37$
$N_{B}=M g \cos 37$
$M_{A}=2$
$M_{B}=4$
$N_{A B}=4 g \times \frac{5}{3}=\frac{12 g}{5}$
$\frac{12 g}{5}+2 g \times \frac{3}{5}=N_{\text {wall}} \frac{4}{5}$
$\frac{18 g}{5}=N_{w a l l} \times \frac{4}{5}$
$N_{\text {wall }}=\frac{9 g}{2}=45$
$M_{A} g \cos 37+N_{\text {wall }} \cos 53=N_{A}$
$N_{A B}+M_{A} g \sin 37=N_{\text {wall }} \sin 53$
$N_{A B}=M g \sin 37$
$N_{B}=M g \cos 37$
$M_{A}=2$
$M_{B}=4$
$N_{A B}=4 g \times \frac{5}{3}=\frac{12 g}{5}$
$\frac{12 g}{5}+2 g \times \frac{3}{5}=N_{\text {wall}} \frac{4}{5}$
$\frac{18 g}{5}=N_{w a l l} \times \frac{4}{5}$
$N_{\text {wall }}=\frac{9 g}{2}=45$
Standard 11
Physics
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