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 450 N at an angle of 15° with the centerline of the canal. What is the net force on the barge?
Fn = 2*450*cos 15 = 869.3 N.
To find the net force on the barge, we need to consider both the magnitude and direction of the forces exerted by the two horses.
The force exerted by each horse can be split into two components: one parallel to the centerline of the canal (the horizontal component) and one perpendicular to the centerline (the vertical component). The vertical components of the forces cancel each other out because the horses are walking along a straight-line section of the canal.
However, the horizontal components of the forces contribute to the net force acting on the barge. To calculate the horizontal component of the force exerted by each horse, we can use the following formula:
F_horizontal = F * cos(theta)
where F is the magnitude of the force and theta is the angle between the force and the centerline of the canal.
In this case, the magnitude of the force exerted by each horse is 450 N, and the angle is given as 15°. Plugging these values into the formula, we get:
F_horizontal = 450 N * cos(15°)
Calculating this expression, we find that the horizontal component of the force exerted by each horse is approximately 433.20 N.
Since there are two horses pulling the barge, we can add up the individual horizontal forces to find the net force:
Net Force = 2 * F_horizontal
Substituting the calculated value of F_horizontal, we get:
Net Force = 2 * 433.20 N = 866.40 N
Therefore, the net force on the barge is approximately 866.40 N, acting parallel to the centerline of the canal.