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When block C is in position xC = 0.8 m, its speed is 1.5 m/s to the right. Find the velocity of block A at this instant. Note that the rope runs around the pulley B and a pin attached to block C, as indicated.

Sagot :

Answer:

The answer is "2 m/s".

Explanation:

The triangle from of the right angle:

[tex]\to (x_c-0.8)+(1.5+y_4) +\sqrt{x_c^2 + 1.5^2}= constant[/tex]

Differentiating the above equation:

[tex]\to V_c +V_A+ \frac{X_cV_c}{\sqrt{x_c^2 +1}}=0\\\\\to 1-V_A+ \frac{0.8 \times 1.5}{\sqrt{ 0.8^2+1.5}}=0\\\\[/tex]

[tex]\to V_A= \frac{1.2}{\sqrt{ 0.64+1.5}}+1\\\\[/tex]

        [tex]= \frac{1.2}{ 1.46}+1\\\\= \frac{1.2+ 1.46}{ 1.46}\\\\ = \frac{2.66}{1.46}\\\\= 1.82 \ \frac{m}{s}\\\\= 2 \ \frac{m}{s}[/tex]

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