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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Physics
Chapter
Laws of Motion

BITSAT 2015 · Physics · Laws of Motion

In figure, two blocks are separated by a uniform strut attached to...

In figure, two blocks are separated by a uniform strut attached to each block with frictionless pins. Block AA weighs 400N400 \mathrm{ N}, block BB weighs 300N300 \mathrm{ N}, and the strut ABA B weigh 200N200 \mathrm{ N}. If μ=0.25\mu=0.25 under B,B, determine the minimum coefficient of friction under AA to prevent motion. In figure, two blocks are separated by a uniform strut attached to each block with frictionless pins. Block A weighs

Options

  1. A0.4
  2. B0.2
  3. C0.8
  4. D0.1

Correct answer

A. 0.4

Step-by-step solution

Consider FBD of structure. In figure, two blocks are separated by a uniform strut attached to each block with frictionless pins. Block A weighs
Applying equilibrium equations, Av+Bv=200NAH=BH…A v+B v=200 \mathrm{ N}A_{H}=B_{H} \ldots (ii) From FBD of block BB, In figure, two blocks are separated by a uniform strut attached to each block with frictionless pins. Block A weighs
BH+FBcos⁡60∘−NBsin⁡60∘=0NBcos⁡60∘−BV−300+FBsin⁡60∘=0FB=0.25NBBH−0.74NB=0B_{H}+F_{B} \cos 60^{{}^{\circ}}-N_{B} \sin 60^{{}^{\circ}}=0N_{B} \cos 60^{{}^{\circ}}-B_{V}-300+F_{B} \sin 60^{{}^{\circ}}=0F_{B}=0.25 N_{B}B_{H}-0.74 N_{B}=0 (iii) −BV+0.71NB=300-B_{V}+0.71 N_{B}=300 FBD of block AA In figure, two blocks are separated by a uniform strut attached to each block with frictionless pins. Block A weighs
FA−AH=0NA−AV=400\ldotsFA=μANA∴μANA−AH=0…F_{A}-A_{H}=0N_{A}-A_{V}=400 \quad \ldotsF_{A}=\mu_{A} N_{A}\therefore \mu_{A} N_{A}-A_{H}=0 \quad \ldots on solving the above equation NA=650N,FA=260N,FA=μANA∴μA=260250=0.4N_{A}=650 \mathrm{ N}, F_{A}=260 \mathrm{ N}, F_{A}=\mu_{A} N_{A}\therefore \mu_{A}=\frac{260}{250}=0.4
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