A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on either side of the canal. Each horse pulls with a force of 700 N at an angle of 10° with the centerline of the canal. What is the net force on the barge?
Each horse pulls with a force of 700 N at an angle of 10° with the centerline of the canal.
700N acts at an angle of 10° to the direction of motion of barge and contributes only a part of the force.
F*sin(10)to the opposite side of angle (along lateral direction) and F*cos(10) to the adjacent side of angle (along centerline of canal)
Net Force from 2 horses = 2*[700*cos(10)]
= 1378.73 N
To find the net force on the barge, we need to determine the horizontal components of the forces exerted by the horses.
Each horse exerts a force of 700 N at an angle of 10° with the centerline of the canal. To find the horizontal component of this force, we can use trigonometry.
The horizontal component (F_horizontal) is given by:
F_horizontal = F * cos(angle)
where F is the magnitude of the force and angle is the angle between the force and the horizontal direction.
In this case, F = 700 N and angle = 10°. Plugging these values into the equation, we have:
F_horizontal = 700 N * cos(10°)
Now we can calculate the horizontal components of the forces exerted by both horses. Since there are two horses, we need to sum up their horizontal components.
Total horizontal force = 2 * F_horizontal
Substituting the value of F_horizontal into the equation, we get:
Total horizontal force = 2 * (700 N * cos(10°))
Finally, we can find the net force on the barge by using the fact that the net force is equal to the vector sum of all the forces acting on the barge.
In this case, since we are only considering the horizontal forces, the net force will be equal to the total horizontal force calculated above.
Therefore, the net force on the barge is given by:
Net force = 2 * (700 N * cos(10°))
Now we can compute the numerical value of the net force.