Question

The motor draws in the cable at c with a constant velocity of vc= 4 m/s. The moto draws in the cable at D with a constant acceleration of aD=8 m/s^2. If vD=0 when t=0, determine: a) the time needed for block A to rise 3 m b)the relative velocity of block A with respect to block B when this occurs.

Answer 1

Answer and Explanation:

The answer is attached below

Answer 2

The motions of the blocks are given by the relationship between displacement, velocity and acceleration.

The correct responses are;

• a) The time needed for block A to rise 3 m is approximately 1.225 seconds.

• b) The relative velocity of block A, with respect to block B is approximately 2.899 m/s.

Reasons:

The given parameters are;

$v_C$ = -4 m/s

$a_D$ = 8 m/s²

a) Required:

The time block A needs to rise 3 meter.

Solution:

Please find attached the diagram of the block as obtained from a similar question online.

Length of the cable extending from point D across the pulley at A and back to the level of the pulley at D; L₁ = $\mathbf{S_D}$ + 2·$\mathbf{S_A}$

$\displaystyle Velocity = \mathbf{ \frac{Distance}{Time}}$

At t = 0, we have;

$v_D$ = 0

ΔL₁ = 0

Therefore;

$v_{L1}$ = 0

Which gives;

$\displaystyle 0 = \frac{S_D + 2 \cdot S_A}{t} = \mathbf{ v_D + 2 \cdot v_A}$

$\displaystyle 0 = v_D + 2 \cdot v_A$

$\displaystyle Acceleration = \mathbf{ \frac{Velocity}{Time}}$

Which gives;

$\displaystyle \frac{0}{t} = 0 = \frac{v_D}{t} + \frac{2 \cdot v_A}{t} = a_D + 2 \cdot a_A$

$\displaystyle 0 = a_D + 2 \cdot a_A$

Therefore;

$\displaystyle a_A = \mathbf{ -\frac{a_D}{2}}$

Which gives;

$\displaystyle a_A = -\frac{8 \, m/s^2}{2} = -4 \, m/s^2$

$\displaystyle a_A =-4 \, m/s^2$

From the kinematic equation of motion, s = u·t + 0.5·a·t², where u = 0, a = $a_A$and s = h, we have;

s = 0.5·a·t²

Which gives;

$h = 0.5 \cdot a_A \cdot t^2$

Therefore;

$\displaystyle t = \sqrt{ \frac{h}{0.5 \cdot a_A} } = \mathbf{\sqrt{\frac{2 \cdot h}{a_A} }}$

Plugging in the values, with up direction taken as negative gives;

$\displaystyle t = \sqrt{\frac{2 \times (-3 \, m)}{-4 \, m/s^2} } = \sqrt{1.5 \, s^2} \approx \mathbf{1.225 \, s}$

• The time needed for block A to rise 3 meters is t ≈ 1.225 s

b) Required:

The relative velocity of block A with respect to block B

Solution:

The relative velocity of block A with respect to block B = $v_{AB} = \mathbf{v_A - v_B}$

The length of the cable at B, L₂ = $S_B$ + $(S_B - S_C)$

L₂ = $2 \cdot S_B - S_C$

ΔL₂ = 0, we have;

$\Delta L_2 = \Delta 2 \cdot S_B - \Delta \cdot S_C = 0$

Which gives;

$\displaystyle \frac{\Delta L_2}{t} = \frac{0}{t} = 0 = \frac{2 \cdot \Delta S_B}{t} - \frac{\Delta \cdot S_C}{t} = \mathbf{2 \cdot v_B - v_C}$

• $\displaystyle 0 = 2 \cdot v_B - v_C$

Which gives;

$\displaystyle 2 \cdot v_B = v_C$

• $\displaystyle v_B = \mathbf{\frac{1}{2} \cdot v_C}$

Therefore;

• $\displaystyle v_B = \frac{1}{2} \times 4 \ m/s = 2 \ m/s$

$v_A = a_A \times t$

$a_A = 4 \ m/s^2$

$t = \sqrt{1.5 \ s^2} \approx \mathbf{ 1.225 \ s}$

Which gives

$v_A = 4 \ m/s^2 \times \sqrt{1.5 \ s^2}$

$v_A - v_B = v_{AB} = 4 \ m/s^2 \times \sqrt{1.5 \ s^2} -2 \ m/s \approx \mathbf{ 2.899 \ m/s}$

• The relative velocity of block A with respect to block B, $v_{AB}$ ≈ 2.899 m/s

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