A block slides at constant speed down a ramp while acted on by three forces: its weight, the normal force, and kinetic friction.
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Answer:
The tensions in is approximately 4,934.2 lb and the tension in
is approximately 6,035.7 lb
Explanation:
The given information are;
The angle formed by the two rope segments are;
The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°
The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°
Therefore, we have;
The tension in rope segment BC =
The tension in rope segment BD =
The tension in rope segment AB = = Pulling force of tugboat = 1200 lb
By resolution of forces acting along the line A_F gives;
× cos(26.0°) +
× cos(21.0°) =
= 1200 lb
× cos(26.0°) +
× cos(21.0°) = 1200 lb............(1)
Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;
× sin(26.0°) +
× sin(21.0°) = 0...........................(2)
Which gives;
× sin(26.0°) = -
× sin(21.0°)
= -
× sin(21.0°)/(sin(26.0°)) ≈ -
× 0.8175
Substituting the value of, , in equation (1), gives;
- × 0.8175 × cos(26.0°) +
× cos(21.0°) = 1200 lb
- × 0.7348 +
×0.9336 = 1200 lb
×0.1988 = 1200 lb
≈ 1200 lb/0.1988 = 6,035.6938 lb
≈ 6,035.6938 lb
≈ -
× 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb
≈ -4934.1733 lb
From which we have;
The tensions in ≈ -4934.2 lb and
≈ 6,035.7 lb.