In the given figure, All surface are smooth & $g = 10 m / s^{2}$ . The force F required to keep 4 kg in equilibria is (Write upto two digits after the decimal point)

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Solution

Free body diagram of the top block is,
$N_{24} cos 53 = 20 \Rightarrow N_{24} = \frac{100}{3}$

Free body diagram of the middle block is,

$N_{34} cos 37 = N_{24} cos 53 + m_{4} g$
$N_{34} \times \frac{4}{5} = \frac{100}{3} \times \frac{3}{5} + 40 \Rightarrow N_{34} = 75 N$
$∴ F = N_{34} sin 37 + N_{24} sin 53 = 71.67 N$

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In the given figure, All surface are smooth & g=10 m/s2. The force F required to keep 4 kg in equilibria is (Write upto two digits after the decimal point)

Free body diagram of the top block is, N24 cos 53=20⇒N24=1003 Free body diagram of the middle block is, N34 cos 37=N24cos 53+m4g N34×45=1003×35+40⇒N34=75N ∴F=N34sin 37+N24sin 53=71.67 N

In the given figure, All surface are smooth & $g = 10 m / s^{2}$ . The force F required to keep 4 kg in equilibria is (Write upto two digits after the decimal point)