A 1 kg mass is suspended from the ceiling by a rope of length 4 m . A horizontal force ' F ' is applied at the mid point of the rope so that the rope makes an angle of 4 5 ∘ with respect to the vertical axis as shown in figure. The magnitude of F is : (Assume that the system is in equilibrium and g = 10 m / s 2 )

T 1 sin 4 5 ∘ = F T 1 cos 4 5 ∘ = T 2 = 1 × g ∴ tan 4 5 ∘ = g F ∴ F = 10 N

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PhysicsNLMFBD & EquilibriumEasy2 minPYQ_2024

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A $1 kg$ mass is suspended from the ceiling by a rope of length $4 m$ . A horizontal force ' $F$ ' is applied at the mid point of the rope so that the rope makes an angle of $4 5^{\circ}$ with respect to the vertical axis as shown in figure. The magnitude of $F$ is : (Assume that the system is in equilibrium and $g = 10 m / s^{2}$ )

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Answer:

Solution:

$T_{1} sin 4 5^{\circ} = F T_{1} cos 4 5^{\circ} = T_{2} = 1 \times g ∴ tan 4 5^{\circ} = \frac{F}{g} ∴ F = 10 N$

Stream:JEESubject:PhysicsTopic:NLMSubtopic:FBD & Equilibrium

⏱ 2mℹ️ Source: PYQ_2024

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