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Include the solution of the problems with the whole procedure clear, complete and organized.
Look for the tension in the rope C
2. Determine the acceleration and tension in string(rope) C, from the following figure, once the masses are released from rest.
1.2 kg 3.2 kg Copyrigh0 2008 Parson Education Ine
Solution
Here the acceleration in both block will be same because there is no change in length of the string during motion
tension on block A(1.2Kg ) and tension on Block B( 3.2Kg ) will be same ..
suppose tension is T then tension in string C will be T+T =2T
suppose acceleration is \'a\' on block A in Upward direction then eq will be
T- 1.2g =1.2a ........(1)
same accelaration on block B but in downward direction eq will be
3.2g-T= 3.2a ............(2)
we get 3.2g - 1.2g = ( 1.2+3.2) a
by solving it a= 4.45 m/s^2
now put a value in eq (1) ...T= 17.11 N
and tension in C will ..= 2T= 34.22 N
.

Include the solution of the problems with the whole procedure clear, complete and organized.
Look for the tension in the rope C
2. Determine the acceleration and tension in string(rope) C, from the following figure, once the masses are released from rest.
1.2 kg 3.2 kg Copyrigh0 2008 Parson Education Ine
Solution
Here the acceleration in both block will be same because there is no change in length of the string during motion
tension on block A(1.2Kg ) and tension on Block B( 3.2Kg ) will be same ..
suppose tension is T then tension in string C will be T+T =2T
suppose acceleration is \'a\' on block A in Upward direction then eq will be
T- 1.2g =1.2a ........(1)
same accelaration on block B but in downward direction eq will be
3.2g-T= 3.2a ............(2)
we get 3.2g - 1.2g = ( 1.2+3.2) a
by solving it a= 4.45 m/s^2
now put a value in eq (1) ...T= 17.11 N
and tension in C will ..= 2T= 34.22 N
.

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