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Two blocks of mass m=5kg and M=10 kg are connected by a string passing over a pulley B as shown. Another string connects the centre of the pulley B to the floor and passes over another pulley A as shown. An upward force F is applied at the centre of pulley A. Both the pulleys are massless. Find the acceleration of both the blocks if 1) F=100N.  2)F=500N
question image
Asked by sakethrockzz007 | 14 Aug, 2023, 16:30: PM
answered-by-expert Expert Answer
Figure shows the free body diagrams of pulleys and blocks. F is applied force on pulley-A .
 
TA and TB are tension forces along the cables connected to pulley.
 
(mg) is weight  of block of mass m . (Mg) is weight of block of mass M.
 
In pulley-A , since pulley's mass is negligible , Tension forces TA balances the applied force.
 
Hence , we have,  begin mathsize 14px style F space equals space 2 space T subscript A end style  
begin mathsize 14px style T subscript A space equals space F over 2 end style . . . . . . . . . . . . . . . .  (1)
In pulley-B , we have
 
begin mathsize 14px style 2 space T subscript B space equals space T subscript A end style
 
begin mathsize 14px style T subscript B space equals space 1 half space T subscript A space end subscript equals space 1 half cross times F over 2 space equals space F over 4 end style
If applied force F = 100 N , then TB = 25 N , block of mass 5 kg moves downward with acceleration am 
 
begin mathsize 14px style m g space minus space T subscript B space equals space m space a subscript m end style
begin mathsize 14px style a subscript m space equals space g space minus space T subscript B over m end style
begin mathsize 14px style a subscript m space equals space left parenthesis space 9.8 space minus space 25 over 5 right parenthesis space m divided by s squared space equals space 3.8 space m divided by s squared end style
---------------------
 
Let aM be the acceleration of block of mass M
 
begin mathsize 14px style M space g space minus space T subscript B space equals space M space a subscript M end style
 
Hence acceleration aM is
 
begin mathsize 14px style a subscript M space equals space g space minus space T subscript B over M end style
begin mathsize 14px style a subscript M space equals space left parenthesis space 9.8 space minus space 25 over 10 right parenthesis space m divided by s squared space equals space 7.3 space m divided by s squared end style
--------------------------------------------------------------------
 
If applied force F = 500 N , then TB = 125 N .
 
Block of mass 5 kg moves upward with acceleration am .
 
begin mathsize 14px style m space a subscript m space equals space T subscript B space minus space m g space end style
begin mathsize 14px style a subscript m space equals space T subscript B over m minus space g end style
begin mathsize 14px style m space equals space left parenthesis space 125 over 5 space minus space 9.8 space right parenthesis space m divided by s squared equals 15.2 space m divided by s squared space end style
--------------------------
 
If aM is the acceleration of block of mass M , then we have
 
begin mathsize 14px style M space a subscript M space equals space T subscript B space minus space M g end style
begin mathsize 14px style a subscript M space equals space T subscript B over M minus space g space equals space left parenthesis space 125 over 10 minus 9.8 space right parenthesis space m divided by s to the power of 2 to the power of space end exponent space equals 2.7 space m divided by s squared space end style

 
Answered by Thiyagarajan K | 14 Aug, 2023, 23:26: PM

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