Question
1) In the first figure, a crate of mass m = 210 kg is pushed atconstant speed up thefrictionless ramp (=26.0°) by a horizontal force F.
(ii) the magnitude of the force exerted by the ramp on the crate.
2) In earlier days, horses pulled barges down canals in the manner shown in the second figure. Suppose that the horse pulls on the rope with a force of 7900 N at an angle of 18° to the direction of motion of the barge, which is headed straight
along the canal. The mass of the barge is 10000 kg, and its acceleration is 0.10 m/s2.
What are the magnitude and the direction of the force on the barge from the water?
Answer
1)
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 θ ... ( 1 ) and
mg = N cos θ ... ( 2 )
Dividing eqn. ( 1 ) by ( 2 ),
F = mg tan θ = 2058 tan (26.0°) = 1004 N
Putting this value of F in eqn. ( 1 ),
N = F / sin θ = 1004 / sin (26.0°) = 2290 N
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
= √[ (6513)^2 + (2441)^2 ]
= 6955 N
If θ is the angle made by this force with the horizontal, then
θ = - arctan (2441) / (6513) = - 20.5°
No comments:
Post a Comment