Find Max Weight of Block A to Keep Block B Stationary

[bob1182006]

[Solved] Tension Help

         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.

 

[learningphysics]

Look at the knot joining the two ropes..

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

 

[m2003]

To find the maximum weight of block A that will keep block B stationary, we need to consider the forces acting on both blocks. For block B, the maximum static friction force (f_smax) is equal to the tension force (T_1) in the string holding it in place. This can be represented by the equation f_{smax}=T_1=\mu_{smax}m_Bg, where \mu_{smax} is the maximum coefficient of static friction and m_B is the mass of block B.

For block A, there are two tension forces acting on it - T_1 from the string holding it in place and T_2 from the string connecting it to block B. In order for block A to remain stationary, these two forces must be equal and opposite, so we can write the equation T_1=T_2\cos41.

To solve for T_2, we need to take into account the weight of block A. This weight creates an additional tension force in the string holding up block A, which we can represent as T_3=m_Ag. Now, we can set up an equation for the forces acting on block A: T_1-T_2\cos41-T_3=0.

Substituting in the values we know, we get T_1-T_2\cos41-m_Ag=0. We can then rearrange this equation to solve for T_2: T_2=\frac{T_1-m_Ag}{\cos41}.

Finally, to find the maximum weight of block A, we need to plug in the maximum tension force (T_1) for block B, which we calculated earlier as \mu_{smax}m_Bg, and the maximum angle of 41 degrees for \cos41. This gives us the equation T_2=\frac{\mu_{smax}m_Bg-m_Ag}{\cos41}.

Solving for m_A, we get m_A=\frac{T_2\cos41}{g}+\frac{m_B\mu_{smax}}{g}. Plugging in the values of T_2=155N, \cos41=0.759, m_B=712N, and \mu_{smax}=0.25, we get a maximum weight of block A as 155N, which matches our previous calculation.

Therefore, the maximum