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Question

Three masses of 2 kg,10 kg and 5 kg are connected to one another and held as shown in figure. If coefficient of friction =0.1 and g=10 m/s2.
Determine:
Value of T1
Value of T2
Value of acceleration
1078396_79eac6e3c31944cd930866bc1250ac08.png

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Solution

Given that,

Mass m1=2kg

Mass m2=10kg

Mass m3=10kg

Friction coefficient μ=0.1

Let T1 and T2 be the tensions in the strings and a be the acceleration towards right

Then

Balancing condition

T2=ma+T1+μmg.....(I)

Now, for T2

T2=mgma....(II)

Now, for T1

T1=m(g+a)....(III)

Now put the value of T1 and T2 in equation (I)

2(10+a)+0.1×10×10+10a=505a

20+2a+10+10a+5a50=0

a=2017

a=1.17 m/s2

Now, put the value of a in equation (II) and (III)

T1=2(10+2017)

T1=23.5N

Now,

T2=5(102017)

T2=44.11N

Hence, the value of acceleration, T1 and T2 is 1.17m/s2 ,23.5N and 44.1N


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