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PhysicsNLMFBD & EquilibriumMedium2 minQB

PhysicsMediuminteger

If the reading of spring balance as shown in figure is 2F newton then value of F is . (Assume that mass M is in equilibrium )

Answer:

Solution:

Normal on the 2.5 kg block: $N = m g cos 3 7^{\circ} = 2.5 \times 10 \times \frac{4}{5} = 20$ N

Spring balance reading: $R = N cos 5 3^{\circ} = 20 \times \frac{3}{5} = 12$ N

$2 F = 12$ , so $F = 6$

Subject:PhysicsTopic:NLMSubtopic:FBD & Equilibrium

⏱ 2mℹ️ Source: QB

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