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Question

# In the figure shown below, a horizontal force F is applied on 5 kg block towards left. If the coefficient of friction between the surfaces are 0.8 and 0.8 as shown in the figure., find the value of tension in the rope and force required just to slide the 5 kg block under the 10 kg block. Take g=10 m/s2.

A
T=50 N,F=120 N
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B
T=62.5 N,F=140 N
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C
T=75 N,F=160 N
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D
T=87.5 N,F=140 N
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Solution

## The correct option is B T=62.5 N,F=140 NThe FBDs of the blocks are as shown~. As per the given question, the block has to just slide which means the condition of friction will be limiting. From the FBD of 10 kg block, we have ∑Fy=0 ⇒ N1=10g−Tsin37∘ ∑Fx=0 ⇒ f1=Tcos37∘−−−−(1) In limiting case, f1=μ1N1=0.8×(10g−Tsin37∘) ⇒ f1=0.8×(100−3T5) ⇒ f1=(80−12T25)−−−−(2) From (1) & (2), we get 4T5=80−12T25⇒ 4T5+12T25=80 ∴ T=62.5 N Now, from the FBD of 5 kg block, we have at equilibrium F=f1+f2−−−−(3) and N2=N1+5g=10g−Tsin37∘+5g ⇒ N2=15g−62.5×35 ⇒ N2=150−37.5 ⇒ N2=112.5 N In limiting case, f2=μ2N2=0.8×112.5=90 N And, from eq. (2), f1=(80−12T25)=(80−12×62.525)=50 N On putting the values of f1 and f2 in eq. (3), we get F=f1+f2=50+90=140 N

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