5/25 At the instant under consideration, the hydraulic cylinder AB has a length L = 0.75 m, and this length is momentarily increasing at a constant rate of 0.2 m/s. If vA = 0.6 m/s and θ = 35°, determine the velocity of slider B.

The image of the question is missing, so i have attached it

Velocity of slider B; = - 0.176 m/s

Explanation:

We are given;

Length of (AB) = 0.75 m

Rate of increase of length; (AB)' = 0.2 m/s

vA = 0.6 m/s

θ = 35°

We want to find vB;

Looking at the image attached, we can use the trigonometric ratio to find OA

Thus;

Sin θ = (OA)/(AB)

So, Sin 35° = (OA)/(AB)

(OA) = (AB)Sin 35°

(OA) = 0.75•Sin 35°

(OA) = 0.75•0.5736

(OA) = 0.43 m

Also, we can use the same system to find (OB)

Thus;

Cos θ = (OB)/(AB)

Cos 35° = (OB)/(AB)

(OB) = (AB)Cos 35°

(OB) = 0.75•Cos 35°

(OB) = 0.75•0.8192

(OB) = 0.6144 m

We apply Pythagoras' theorem as follows

(AB)² = (OA)² + (OB)²

We derive the equation;

2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB

Divide through by 2 to give;

(AB)*(AB)' = (OA)*vA + (OB)*vB

vB = ((AB)*(AB)' - (OA)*vA) / (OB)

We now have ;

vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s)/(0.614 m)

vB = - 0.176 m/s

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explanation:

it was a fall down when it happens

explanation: