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ans for first part wud be 150N,,,,,,,,,,,,,,,,,,just wait for 5min ..........m writing the whole solution
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value of g at poles is slightly higher .....................so time period of pendulum will decrease
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so for mass M block the equation will be
T- Mg sin30 = M*2a
for mass 2M
2Mg-2T=2M*a
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u have got constraint equation exactly right...........just proceed after making the free body diagram of both bodies
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see on the pulley(which is attached to clamp).............the string is passing such that its two part makes 90degree with each other............since both the parts of string have tension T............so by vector addition force on pulley will be (sq rt.2 )T...........same is the force exerted by clamp on pulley.....
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first u should know the constraint equations.............which means if the accelration of 2M is a.................then acceleration of M is 2a
for 2M force equation is
2T = 2M* a
and for mass M
Mg - T = M*2a
find T and a
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accleration is 2g/3 of mass M....................and of 2M is g/3
T=Mg/3.........and the force on clamp is F =(sq.rt.2)*Mg/3
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since whole block is of 3kg and 30cm.................so 10cm will weigh 1kg and 20cm will weigh 2kg
now the force equation will be
32 - F = 2*a
F-20 = 1*a
a is the accleration of whole block..............and F is the contact force b/w two parts.......
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answer is 24 N...............WAIT M WRITING THE WHOLE SOLUTION
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good........ur procedure is totally correct
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i think the answer is option A
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relative velocty of A wrt B = 30-20 =10m/s
now the acc = -2 m/s^2 as the direction of both acc and rel velocity is diffrent
now to avoid collision they should just meet.........so......
0= 10^2 - 2*2*d
d= 100/4=25m
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bcause it is AB2 type and crystallized in cubic close packing
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after 1 sec the upward speed gained by the stone will be 2m/s
and it will reach the height = 1/2 * 2 *(1^2)=1m
1= -2t +1/2 *10*t^2 here i have taken upward direction as negative
5*t^2 - 2t -1
find t by this
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