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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 N
The 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=10gTsin37
Fx=0
f1=Tcos37(1)
In limiting case,
f1=μ1N1=0.8×(10gTsin37)
f1=0.8×(1003T5)
f1=(8012T25)(2)

From (1) & (2), we get
4T5=8012T25 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=10gTsin37+5g
N2=15g62.5×35
N2=15037.5
N2=112.5 N
In limiting case,
f2=μ2N2=0.8×112.5=90 N

And, from eq. (2),
f1=(8012T25)=(8012×62.525)=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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