Problem 5.104 Part A The 4.00-kg block in the figure (Figure 1) is attached to a
ID: 1588689 • Letter: P
Question
Problem 5.104 Part A The 4.00-kg block in the figure (Figure 1) is attached to a vertical rod by means of two strings. When the system rotates about the axis of the rod, the strings are extended as shown in the diagram and the tension in the upper string is 75.0 N What is the tension in the lower cord? Figure 1 of 1 Submit My Answers Give Up Incorrect; Try Again; no points deducted 1.25 m Part B 2.00 m 4.00 kg How many revolutions per minute does the system make? 1.25 m rev/min Submit My Answers Give UpExplanation / Answer
Geometry gives us that the block is 0.75 m from the rod.
The strings are = arccos(1/1.25) = 37º from the rod.
A) Vertical equilibrium gives us that Tu*cos37 = Tb*cos37 + 4kg*9.8m/s²
where Tu, Tb are the tensions in the upper, lower strings, respectively.
Then 0.8Tu = 0.8Tb + 39.2N
and Tu = Tb + 49N (after dividing through by 0.8)
Tb = Tu - 49N = 75N - 49N = 26 N
B) Horizontal equilibrium gives us that m²r = (Tu + Tb)sin
4kg * ² * 0.75m = 119N * sin37º
² = 23.8 rad/s²
= 4.9rad/s * 60s/min * 1rev/2rads = 46.6 rpm
C) When the lower cord goes slack, Tb = 0 and = 37º still, so
horizontal equil: m²r = Tu*sin37 3² = 0.6Tu
vertical equil: mg = Tu*cos37 39.2 = 0.8Tu Tu = 49
plug that into horiz:
² = 0.6*49 / 3 = 9.8 rad/s²
= 3.13rad/s * 60s/min * 1rev/2rads = 29.9 rpm