In the figure, a horse pulls a barge along a canal by means of a rope. The force

Question

In the figure, a horse pulls a barge along a canal by means of a rope. The force on the barge from the rope has a magnitude of 7770 N and is at the angle ? = 21

Explanation / Answer

follow this Three forces act on the crate: i) its wt. (210)*(9.8) = 2058 N acting downwards, ii) Horizontal force, F and iii) Reactive force, N, from the ramp on the crate perpendicular to the inclined plane. Since mass is pushed at constant speed, there is no net force acting on the crate which means that these three forces balance each other. Balancing horizontal and vertical component of forces, F = Nsin ? ... ( 1 ) and mg = N cos ? ... ( 2 ) Dividing eqn. ( 1 ) by ( 2 ), F = mg tan ? = 2058 tan (26.0°) = 1004 N Putting this value of F in eqn. ( 1 ), N = F / sin ? = 1004 / sin (26.0°) = 2290 N I shall come back to answer next question after 2 hours. 2) Two force act on the barge in the horizontal direction: i) Horizontal component of tension, T in the rope = 7900 cos 18° = 7513 N and ii) Opposing drag force from water on the barge = F Under the effect of these two forces, the barge has acceleration of 0.10 m/s^2. So, (10000) * (0.10) = 7513 - F => F = 6513 N There is another vertical reactive force, F', on the barge from water which is in the downward direction opposite to the upward component of tension T in the rope = T sin 18° = 7900 * sin 18° = 2441 N Net force from water is the resultant of F and F' Magnitude of net force from water = v[ (6513)^2 + (2441)^2 ] = 6955 N If ? is the angle made by this force with the horizontal, then ? = - arctan (2441) / (6513) = - 20.5°

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