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Question

# In the figure shown below, a horizontal force F is applied on 4 kg block towards left. If the coefficient of friction between the surfaces are 0.5 and 0.4 as shown in the figure. The value of Tension in the rope and force required just to slide the 4 kg block under 2 kg block is (rope is massless and inextensible)

A
T=90.9 N and F=29 N
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B
T=9.09 N and F=14.5 N
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C
T=9.09 N and F=29 N
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D
T=90.9 N and F=14.5 N
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Solution

## The correct option is C T=9.09 N and F=29 NThe FBDs of the blocks are as shown below R1 and R2 are the normal forces acting on the respective blocks. The condition is just to slide, it means in the question it is talking about limiting condition. So, from FBD of 2 kg we have, R1+Tsin37∘=2g=20 R1=20−3T5 ... (1) f1=Tcos37∘ ⇒ f1=4T5 ... (2) ∵ f1=μR1 ⇒ f1=0.5(20−3T5)=10−3T10... (3) From (3) and (2) we get 4T5=10−3T10 4T5+3T10=10 11T10=10⇒T=10011 N ... (4) From (4) in (1) we get R1=20−3×1005×11 =20−5.454 R1=14.546 N Now, from the FBD of 4 kg we get, At equilibrium, R2=R1+4g=14.546+40=54.546 N F=f1+f2 ∵f2=μ2R2=0.4×54.546=21.8 N and from equation (3) f1=10−3×10010×11=7.27 ⇒F=21.8+7.3=29.1 N

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