In the arrangement shown above, the block of mass m = 2 kg lies on the wedge of mass M = 8 kg . Find the initial acceleration of the wedge if the surfaces are smooth, pulley and strings are massless.
If the magnitude of acceleration is in the form c a b , where a , b and c are positive integers , with a , c coprime and b square-free, submit your answer as a + b + c .
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I don't get it b=a/2. Could you tell me why?
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Pulley P 1 is moving with acceleration a towards the right since it is fixed on the wedge. So the component of this acceleration towards P 2 which is a cos 6 0 gives us b = 2 a .
Can you please help me ? Why is the acceleration of the object =3/2a-acos60
Not to be picky or anything but in the problem, it's stated clearly that we have to find the initial acceleration. The initial acceleration means we can stop worrying about time and think about the instantaneous moment after the the block of mass m is released. Then to stop the mass m from falling mass M has to provide a normal force (normal to the incline of the wedge) to stop it. Moreover, the rope is basically a connector ( massless rope) so the tangetial component of the weight on mass m along the incline is transmitted through the rope and used to pull the wedge to the right. And if you solved it this way, you never get 56. I dare you to prove my explaination wrong and tell me how yours is correct. And we're studying science here. Give me evidence.
i don't understand why you have taken b between p1 and p2
Can you please explain how you found b=a/2 ? As b is on the hypotenuse side and a is on base side, should that not mean b>a for the right triangle? I think for the initial movement a=bcos60 which gives a=b/2. which gives b=2a. Please correct me if I am wrong. Thanks.
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H = l 2 + b 2 now differentiate the equation w. r. t time you will get the required result and remember here only one of l or b is changing with time.
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I Did this way
Consider The wedge to be moving with an acceleration a1 towards the right.
Call the tension in the string throughout to be T and normal reaction with wedge and block to be N.
So we can write T+Nsin(60) = Ma1
Next to analyse motion of block lets go into the frame of wedge (Non Inertial frame)
Call the acceleration of block along wedge to be a2 in frame of wedge
Writing equations for the block along wedge and perpendicular to wedge (Do not forget to apply Pseudo force as we are working in non inertial frame of reference)
N+ma1sin(60) = mgcos(60)
T-ma1cos60 -mgsin(30) = ma2
Notice that acceleration of block along the wedge wrt ground is a2 + a1/2
Applying Virtual work method for constraints we obtain
a2+3a1/2 = 0
Solving above system of equations we get required answer
A Really nice problem on Laws of motion :)
A reaIly very nice solution.I did somewhat like this but instead of applying pseudo force,i have found the relationship between the accelaration of wedge and accelaration of block w.r.t. wedge by applying work done by tension=0 and luckily due to the data given,both acceleration came the same.So by applying newtons laws from ground frame for the block with suitable coordinate axis.the ans comes the same.
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Ohh nice!. actually i substituted all values in the end!
Dear Mr. Shubhendra Singh, the correct "string constraint" equation for the accelerations (which you can easily get by equating the time-derivative of the length of the string to zero) lookes different from what you used, although it would be correct in case of velocities. The origin of the correction is in the change of P1-P2 side.
Did the same! In fact, even my variables were the same :p
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a : acceleration of the wedge
b : acceleration by which P 1 P 2 is decreasing.
T : Tension in the string
(All accelerations shown in the image are in ground frame)
By applying string constraint we can say that b = 2 a .
Now use the approach of system of masses for getting the equations.
⇒ 2 g sin 6 0 − T = 2 ( 2 3 a − a cos 6 0 )
⇒ 2 3 T = 8 a + 2 ( a − 2 3 a cos 6 0 )
Solving these equations we get T = 3 1 7 a and a = 2 3 3 0 3
So the answer is 3 0 + 2 3 + 3 = 5 6