Question

The tractor A is used to hoist the bale B with the pulley arrangement shown. If A has a forward velocityv_A, determine an expression for the upward velocity v_B of the bale in terms of x.

• A. $\frac { \mathrm { v } _ { \mathrm { A } } \mathrm { x } } { \mathrm { h } }$
• B. $\frac { 1 } { 2 } \frac { \mathrm { v } _ { \mathrm { A } } \mathrm { x } } { \sqrt { \mathrm { h } ^ { 2 } + \mathrm { x } ^ { 2 } } }$
• C. $\frac { 1 } { 2 } \frac { \mathrm { v } _ { \mathrm { A } } \mathrm { h } } { \sqrt { \mathrm { h } ^ { 2 } + \mathrm { x } ^ { 2 } } }$
• D. $\frac { v _ { \mathrm { A } } \mathrm { h } } { \mathrm { x } }$

Answer 1

Answer: (B)

$\frac { 1 } { 2 } \frac { \mathrm { v } _ { \mathrm { A } } \mathrm { x } } { \sqrt { \mathrm { h } ^ { 2 } + \mathrm { x } ^ { 2 } } }$

Answer 2

We designate the position of the tractor by the coordinate x and the position of the bale by the coordinate y, both measured from a fixed reference. The total constant length of the cable is

L = 2(h – y) + l = 2 (h – y) + $\sqrt { h ^ { 2 } + x ^ { 2 } }$

Differentiation with time yields

0 = – $2 \dot { \mathrm { y } }$ + $\frac { \mathrm { x } \dot { \mathrm { x } } } { \sqrt { \mathrm { h } ^ { 2 } + \mathrm { x } ^ { 2 } } }$

Substituting v_A = $\dot { \mathrm { X } }$ and v_B = $\dot { \mathrm { y } }$ gives

v_B = $\frac { 1 } { 2 }$ $\frac { \mathrm { xv } _ { \mathrm { A } } } { \sqrt { \mathrm { h } ^ { 2 } + \mathrm { x } ^ { 2 } } }$