Please show all the small steps in your process, even if it seems like the way y

Question

Please show all the small steps in your process, even if it seems like the way you got from one step to another is obvious. I don't know how to set it up, so please include why you have set up the problem the way that you have (as in, how do you know to do it the way that you do?). Thank you!

Block B in (Figure 1) rests on a surface for which the static and kinetic coefficients of friction are 0.65 and 0.40, respectively. The ropes are massless.

What is the maximum mass of block A for which the system remains in static equilibrium?

Problem 7.7 - Enhanced - with Feedback Part A Block B in (Figure 1) rests on a surface for which the static and kinetic coefficients of friction are 0.65 and 0.40, respectively. The ropes are massless. What is the maximum mass of block A for which the system remains in static equilibrium? Express your answer with the appropriate units. You may want to review (-pages 165-168) 0? | Figure 1 of 1 mx= 1 Value Units Submit My Answers Give Up 20 kg 45° 90°

Explanation / Answer

Solution-

Block B will slip when the tension on the horizontal rope exceeds
20 * 9.8 * 0.65 = 127.4 N
Therefore the maximum allowable tension on the horizontal rope is 127.4 N.

Now this 117.6 Newton force to the left must be equal to some force to the right in order for the system of ropes to be in equilibrium, this is the horizontal component of the tension on the angled segment of rope.

The horizontal component of the tension on the angled segment of rope = 127.4 N

Now since the rope is angled at 45 degrees, the vertical component is also equal to 127.4 N

which in turn is equal (but opposite) to the tension on the vertical rope.

So, mass A equals
127.4 / 9.8

= 13.0 kg

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