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23 Oct 2007 15:43:49 IST

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884

solve it for salutes

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consider a 3 block system consisiting of blocks A ,B &C OF MASSES 5 kg 10 &15 kg with 5 kg placed over 10 kg &10 kg over 15 kg block. the coefficient of friction betn 5 kg block & 10 kg block &betn 10 kg block &15 kg block is 0.2

the coeff. of friction betn 15 kg block & table surface is 0.5 now a force of 50 newton is applied on 5 kg block & 100N force on 10 kg block aind the accln of each block

KINDLY GIVE A DETAILED SOLN EXPLAINING THE DIR. OF FRICTION FORCE IN EACH CASE

Comments (4)

Vivek Sarda

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Joined: 21 Oct 2007

Posts: 271

23 Oct 2007 15:48:47 IST

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for the top body only friction is backward direction
on the second body friction by the upper body froward n by lower body backward
on the third body friction by ground backward n by the upper body forward
then write newtons 2nd law u ll getthe solution
if u r still nt satisfied nudge book me

divya jyothi

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Joined: 24 Oct 2007

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24 Oct 2007 14:07:42 IST

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Supposing that the two forces are acting in the same direction:- The first block hastwo forces acting on it i.e 50N&max fr=10N which act in opposite directions.So resultant force acting is 40N .so acc is 40/5=8m/s^.For the second block,100N&the previous frmax i.e10N, frmax between10kg,15kg which is 30N are the forces acting.The two forces:100N,10N act opposite to the 30N force.Now the resultent force is 80N.So the acc.of 10kg block=80/10=8m/s The third block,is acted upon by two forces-one of the pairing forces between 10kg&15kg=30N,&frmax between 15kg & the table=15Resultent=15N,acc.of 15kg block=15/15=1m/s^

GAURAV

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30 Oct 2007 10:26:54 IST

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hi rose plz also tell me that uptill when the bodies will have a common accl. &wen "ll they separate

Soumen Goswami

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27 Nov 2007 20:23:43 IST

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Assumption: Both the forces are acting in the same direction, say rightwards.
Let the friction force between the 5 and 10 kg blocks be f₁, between 10 and 15 kg block be f₂ and between 15 kg and the ground be f₃.
f_1max = 0.2 x 5 x 10 = 10 N
f_2max = 0.2 x 15 x 10 = 30 N
f_3max = 0.5 x 30 x 10 = 150 N
The upper block can't stay in equilibrium as the applied force is greater than the maximum value of the frictional force. The net rightward force on the second block is 100 + 10 =110 N, which is greater than the maximum value of the frictional force between the 2nd and third block. So the second block too will accelerate. However, the net force on the bottommost block is 30 N rightwards, which is lesser than the maximum frictional force between the 3rd block and the ground. Hence the 3rd block remains at rest.
a₁=(50-10)/5 = 8
a₂=(110-30)/10 = 8
a₃= 0

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