Question

A crane at a construction site is used to lift and move steel beams and place them on the building under construction as needed. The motion of the beam is measured tension in the cable and the time intervals the tension is applied.

For example, consider the following situation:

• The beam is initially suspended, stationary, above the ground. The tension in the cable is T₀ = 2580 N.
• Next, the tension in the cable is T₁ = 2660 N for a time period $LaTeX: \Delta t_1$ = 1.81 seconds.
• Next, the tension in the cable is T₂ = T₀ = 2580 N for a time $LaTeX: \Delta t_2$ = 2.13 seconds.
• Finally, the tension in the cable is T₃ = 2390 N until the steel beam is again stationary.

How far did the steel beam move from its initial position to its final position, where it is again stationary?

Give your answer in meters to at least three significant digits.

Note: Yes, this problem is a combination of forces and motion.

Answer 1

initially tension in cable (when beann is stationery) y 2580 H which should be caual to it's self weight. So, mg= 2580 am= 26319) How for sti=1.81 seconds Tensian who was - 2660 N > 2580 to beam will more upwand direction, let's calculate how much 2660 groa 2660-mg ama 2860-2580=2639 sa=0.304 mife 2 so, sa ut t I at? a fe 0+ * * 0.204*(1-8132 3 48m , asar Si=0,498 m) by this time it must have gained some velocity also leth see how much,

- rautat a ų 0+ 0.2044 1781 alų= 0.550 24 m/s) - # again tension tz į equal to it's own weight mears lazo So, with v velocity i t will cover some distance in time atz za Vix ata a fz = 0.5502442.17 3 5z = 1.172 ml # Now De after Ata time still beam is moving so, tension has to be decreased for stopping it but during this process also beam will move some distance let's calculate how much no 2390 ng-2990= ma arter 2580-2390= 263a ala=0.722 m2 (downward) so, distance travelled before shopping is U² u2_zas 03 (0.590243- 2*0.722 * Sa » [f3= 0,21 m. so, tored Distance covered by beam sa fit fat fg S= 0.448 +1,172 +0.21 » als= 1,8796 m.]

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