Find Max Weight of Block A to Keep Block B Stationary

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The discussion focuses on determining the maximum weight of block A that allows block B to remain stationary on a tabletop. Block B has a weight of 712N and a static friction coefficient of 0.25, which is crucial for calculating the forces involved. The equations of motion for both blocks are analyzed, particularly the relationship between the tensions in the strings and the forces acting on block B. After some calculations and adjustments, the correct maximum weight for block A is found to be approximately 155N. The solution highlights the importance of considering all forces and tensions in the system to arrive at the correct answer.
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[Solved] Tension Help

Code:
         T2 /
---_T1____./ 41*
  B       |
---       |
         |  |
         |A |

A and B are blocks. B is stationary on a tabletop. A is hanging off the edge beind held by 2 strings.

Homework Statement


Block B weighs 712N. The coefficient of static friction between block B and the table is 0.25. Find the maximum weight of block A for which block B will remain at rest.


Homework Equations


F=ma

The Attempt at a Solution



Block B:
f_{smax}-T_1=0
f_{smax}=T_1
f_{smax}=\mu_{smax}m_Bg

Block A
T_1-T_2\cos41=0
T_1=T_2\cos41

I can't figure out what T2 is. I thought at first it would be m_Ag But plugging that in I get mass of Block A is 235.85N but it should be 155N.

Edit: forgot about the tension between block A and the string holding it up which allowed me to get 154.7~155N when rounded.
 
Last edited:
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Look at the knot joining the two ropes..

write the \Sigma{F_y} = 0 equation...
 
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