Physics, 04.12.2020 17:00 hoyanna69

Determine the force P required to maintain the 200-kg engine in the position for which θ = 30°. The diameter of the pulley at B is negligible.

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Answer from: lame3245

The force P required is 1759.22 N

Explanation:

The missing diagram is seen in the first image below.

From the second image, we can see the schematic diagram of the engine hanging over the pulley.

To start with determining the value of the angle ∝;

$tan \ \alpha = \dfrac{CD}{BD}$

where;

BD = AB-AD

Then;

$tan \ \alpha = \dfrac{CD}{AB-AD}$

$\alpha = tan^{-1} \bigg(\dfrac{CD}{AB-AD} \bigg )$

replacing their respective values, where;

CD = 2 sin 30° m, AB = 2m and AD = 2 cos 30° m

$\alpha = tan^{-1} \bigg(\dfrac{2 \ sin \ 30^0}{2-2 \ cos \ 30^0} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{2-1.732} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{0.268} \bigg )$

$\alpha = tan^{-1} \bigg(3.73\bigg )$

$\alpha \simeq 75^0$

From the third diagram attached below:

The tension occurring in the thread BC is equal to force P

$T_{BC} = P$

Using the force equilibrium expression along the horizontal direction.

$\sum F_x = 0\\\\ -T_{AC} \ cos \ 30^0 + Pcos \alpha = 0$

replacing the value of $\alpha \simeq 75^0$

$-T_{AC} \ cos 30^0 + P cos 75^0 = 0$

$P \ cos \ 75^0 = T_{AC} \ cos \ 30^0$

$P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0} \ \ \ - - - (1)$

Along the vertical direction, the force equilibrium equation can be expressed as:

$\sum F_y =0$

$-W + P \ sin \alpha + T_{AC} \ sin \ 30^0 = 0$

$W = P \ sin \ \alpha + T_{AC} \ sin \ 30^0$

replacing $\alpha \simeq 75^0$ and $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$W =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

Also, replacing W for (200 × 9.81) N

$200 \times 9.81 =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

$200 \times 9.81 = T_{AC} \ cos \ 30^0 \ tan \ 75^0 + T_{AC} \ sin \ 30^0$

$1962= T_{AC} \ ( cos \ 30^0 \ tan \ 75^0 + \ sin \ 30^0)$

$1962= T_{AC} \ (0.8660\times 3.732 + 0.5)$

$1962= T_{AC} \ (3.231912 + 0.5)$

$1962= T_{AC} \ (3.731912)$

$T_{AC} = \dfrac{1962}{ \ (3.731912)}$

$T_{AC} = 525.736 \ N$

From $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \times0.866}{0.2588}$

P = 1759.22 N

Thus, the force P required is 1759.22 N

Determine the force P required to maintain the 200-kg engine in the position for which θ = 30°. The

Answer from: Quest

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Answer from: Quest

Part a: using momentum conservation we have: direction x: (15) (7.0) + 0 = (15) (4.2) cos (-20 °) + (10) v₂ (f) cosθ (10) v₂ (f) cosθ = (15) (7.0) - (15) (4.2) cos (-20 °) (eq 1) direction y: 0 + 0 = (15) (4.2) sin (-20 °) + (10) v₂ (f) sinθ (10) v₂ (f) sinθ = - (15) (4.2) sin (-20 °) (eq 2) we divide both equations (1) and (2): left side: (10) v₂ (f) sinθ / (10) v₂ (f) cosθ = tanθ right side: - (15) (4.2) sin (-20 °) / (15) (7.0) - (15) (4.2) cos (-20 °) clearing the angle we have: θ = tan⁻¹ [- (15) (4.2) sin (-20 °) / (15) (7.0) - (15) (4.2) cos (-20 °)] θ = 25 ° we use any of the equations to find the speed. from (2) we have: (10) v₂ (f) sinθ = - (15) (4.2) sin (-20 °) v₂ = - (15) (4.2) sin (-20 °) / (10) sin25 ° answer: v₂ = 5.1m / s note: θ = 25 ° (direction is in the first quadrant, north of east) part b: initial kinetic energy: k (i) = 0.5 (15) (7.0) ² = 370j (rounded) final kinetic energy: ke (f) = 0.5 (15kg) (4.2m / s) ² + 0.5 (10kg) (5.1m / s) ² = 260j (rounded) answer: a difference of 110j, so it was an inelastic collision.

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