A motor vessel tows a small dinghy by a rope which makes an angle of 20 degrees with the horizontal. If the tension in the rope is 15 kgf ;find

a) the force which effectively pulls the dinghy forward.
b)the force which lift its bows out of the water.

horizontal 15cos 20

vertical 15 sin 20

Question

Question

Physics

To find the force which effectively pulls the dinghy forward, we need to analyze the forces acting on the dinghy.

Let's consider the forces in the horizontal and vertical directions separately.

a) Force pulling the dinghy forward:
The horizontal component of the tension force is responsible for pulling the dinghy forward.

To find the horizontal component of the tension, we can use the trigonometric relationship between the angle and the components of the force. The horizontal component can be found using the formula:

Horizontal component of the force = Tension force * cos(angle)

Here, the tension force is given as 15 kgf and the angle is given as 20 degrees.

Horizontal component of the force = 15 kgf * cos(20 degrees)

Calculating this, we find that the force pulling the dinghy forward is approximately 14.35 kgf.

b) Force lifting the bows out of the water:
The vertical component of the tension force is responsible for lifting the bows of the dinghy out of the water.

To find the vertical component of the tension, we can also use the trigonometric relationship between the angle and the components of the force. The vertical component can be found using the formula:

Vertical component of the force = Tension force * sin(angle)

Using the given values:

Vertical component of the force = 15 kgf * sin(20 degrees)

Calculating this, we find that the force lifting the bows of the dinghy out of the water is approximately 5.11 kgf.