Block B in the figure(Figure 1) rests on a surface for which the static and kine
ID: 1325543 • Letter: B
Question
Block B in the figure(Figure 1) rests on a surface for which the static and kinetic coefficients of friction are 0.63 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 to two significant figures and include the appropriate units.
Block B in the figure(Figure 1) rests on a surface for which the static and kinetic coefficients of friction are 0.63 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 to two significant figures and include the appropriate units.Explanation / Answer
If the tension on the horizontal rope exceeds 123.48 Newtons then block B will slip.
20 * 9.8 * 0.63 = 123.48
This is the maximum allowable tension on the horizontal rope.
In order for the system of ropes to be in equilibrium, this 123.48 Newton force to the left must be equal to some force to the right, namely, the horizontal component of the tension on the angled segment of rope.
The horizontal component of the tension on the angled segment of rope equals 123.48 Newtons.
Because the rope is angled at 45 degrees, the vertical component is also equal to 123.48 Newtons.
...which in turn is equal (but opposite) to the tension on the vertical rope.
Therefore, mass A equals 12.6 kg.
123.48 / 9.8 = 12.6
Incidentally, Mass A = Mass B * the coefficient of static friction.
This would not be true, however, if the angled segment were at a different angle.