Question

Years ago the barges that traveled along the canals were dragged by horses. Suppose that the horse is exerting a force of 7900 N at an angle of 18 degrees with the direction of movement of the barge, which is sailing in a straight line through the channe?

Years ago the barges that traveled along the canals were dragged by horses. Suppose that the horse is exerting a force of 7900 N at an angle of 18 degrees with the direction of movement of the barge, which is sailing in a straight line through the channel, the mass of the barge is 9500 kg and its acceleration is 0.12 m / s ^ 2.

1-Calculate the force exerted by the water on the barge
2-Find the direction of the force exerted by the water on the block

Answer 1

`F_"water" = 6824 N" "21^@` to starboard of astern.#

Explanation:

From the mass and acceleration, we can find the net force acting on the barge.

`F_"net" = 9500 kg * 0.12 m / s^2 = 1140 N`
`F_"net"` is a force parallel to the canal since that is the direction of the acceleration.

We need to resolve the 7900 N into the forces parallel with the canal and perpendicular to it.

`F_"par" = 7900 N*cos18^@ = 7513 N`
`F_"perp" = 7900 N*sin18^@ = 2441 N`

`F_"perp"` is a force pointed to the left of the barge's course. Since the barge is moving parallel with the canal, the net force perpendicular to the canal is zero. Therefore the water must be opposing `F_"perp"` with an equal and opposite force: 2441 N perpendicular to the canal to the right of the barge's course.

`F_"net" = 1140 N and F_"par" = 7513 N`. These forces are pointed forward and are parallel with the canal. `F_"net"` must be the sum of `F_"par"` and the force that water applies to the barge parallel with the canal.

`F_"net" = F_"par" + F_"water-par"`

Therefore

`F_"water-par" = F_"net" - F_"par" = 1140 N - 7513 N`

`F_"water-par" = -6373 N`

The net force applied to the barge by the water is the resultant of the water's forces. Draw those forces. The water's 2441 N is perpendicular to the canal to the right of the barge's course. And `F_"water-par" = -6373 N` -- that is perpendicular to the canal to the rear.

We find the magnitude of the resultant using Pythagoras:

`F_"water" = sqrt(2441^2 + 6373^2) N = 6824 N`

The direction of the water's force is `arctan(2441/6373) = 21^@`. From your drawing of the water's 2 component forces, it should be clear that the direction of resultant is mostly to the rear -- slightly (`21^@`) to the right.

I hope this helps,
Steve