Homework SourseYou are an engineer working for a manufacturing company You
You are an engineer working for a manufacturing company. You are designing a mechanism that uses a cable to drag heavy metal blocks a distance of 8.00 m along a ramp that is sloped at 40.0 above the horizontal. The coefficient of kinetic friction between these blocks and the incline is k = 0.350. Each block has a mass of2090 kg . The block will be placed on the bottom of the ramp, the cable will be attached, and the block will then be given just enough of a momentary push to overcome static friction. The block is then to accelerate at a constant rate to move the 8.00 m in 4.20 s. The cable is made of wire rope and is parallel to the ramp surface. The table gives the breaking strength of the cable as a function of its diameter; the safe load tension, which is 20% of the breaking strength; and the mass per meter of the cable:
A.) What is the minimum diameter of the cable that can be used to pull a block up the ramp without exceeding the safe load value of the tension in the cable? Ignore the mass of the cable, and select the diameter from those listed in the table. inches.
B.)
Graphing either SL versus D2 or D2 versus SL gives a straight line. In the graph shown, we chose to plot SL on the vertical axis and D2 on the horizontal axis.
Estimate the safe load value for a cable with diameter 916 in. using the equation for the line of best fit given.
| Cable Diameter (in.) | Breaking Strength (kN) | Safe Load (kN) | Mass per Meter (kg/m) |
| 14 | 24.4 | 4.89 | 0.16 |
| 38 | 54.3 | 10.9 | 0.36 |
| 12 | 95.2 | 19.0 | 0.63 |
| 58 | 149 | 29.7 | 0.98 |
| 34 | 212 | 42.3 | 1.41 |
| 78 | 286 | 57.4 | 1.92 |
| 1 | 372 | 74.3 | 2.50 |
Solution
mass of the load = 2090 kg
when the load is accelerated upward force acting on the load are
1. wieght component down the plane mgSin(40)
2. kienetic friction down the plane 0.35*mgCos(40)
3. tension in the cable up the plane T
if a is the acceleration of the load motion of the load is govenrned by
T- mgSin(40) - 0.35*mgCos(40) = ma
the block moves 8.00 m in 4.2 s
s= at2/2
accelerationof the block up the plane a = sqrt(8*2/4.2*4.2) = 0.95 m/s-s
T = m(a+gSin(40) + 0.35*gCos(40))
= 2090*(0.95 +9.8*Sin(40) +0.35*9.8*Cos(40))
= 20643 N
safe load is 20% hence the cable breaking strength shall be five times the tension
= 5T = 103214 N
we shall use the cable 58 with safe load 29.7 kN and breaking strength 149 kN
b) SL = 74.1kN *(0.916)2 + 0.499kN = 62.67 kN
