A lane mower is purhed with aforce of 50n if the angle between the handle of the mower and the ground is 30

a calculate the magnitude of the force that is pressing the lane mower directly into the ground
b calculate the magnitude of the effective force that moves the mower forward
c why does the lawn mower move foward and not downward into the ground?

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a) To calculate the magnitude of the force pressing the lawn mower into the ground, we need to find the component of the force that acts perpendicular to the ground. This component is given by the equation:

Force perpendicular to ground = Force applied * cos(angle)

Given that the force applied is 50 N and the angle is 30 degrees, we can substitute these values into the equation:

Force perpendicular to ground = 50 N * cos(30 degrees)

Using the cosine of 30 degrees, which is √3/2, we can simplify the expression:

Force perpendicular to ground = 50 N * (√3/2) ≈ 43.3 N

Therefore, the magnitude of the force pressing the lawn mower into the ground is approximately 43.3 N.

b) To calculate the magnitude of the effective force that moves the mower forward, we need to find the component of the force that acts parallel to the ground. This component is given by the equation:

Force parallel to ground = Force applied * sin(angle)

Substituting the values from the previous question:

Force parallel to ground = 50 N * sin(30 degrees)

Using the sine of 30 degrees, which is 1/2, we can simplify the expression:

Force parallel to ground = 50 N * (1/2) = 25 N

Therefore, the magnitude of the effective force that moves the lawn mower forward is 25 N.

c) The lawn mower moves forward and not downward into the ground due to the law of physics known as Newton's second law. According to this law, the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. In this case, the force applied to the lawn mower is greater in the forward direction (25 N) than in the downward direction (approximately 43.3 N). Hence, the lawn mower accelerates forward, resulting in its forward motion.

To answer these questions, we need to understand the concept of vector decomposition and Newton's laws of motion.

a) To calculate the magnitude of the force that is pressing the lawn mower directly into the ground, we need to decompose the 50N force into its vertical and horizontal components. Since the angle between the handle and the ground is given as 30° and the force is being pushed, the vertical component of the force is responsible for pressing the lawn mower into the ground.

By using trigonometry, we can find the vertical component:

Vertical Component = Force × sin(angle)
Vertical Component = 50N × sin(30°)
Vertical Component = 50N × 0.5
Vertical Component = 25N

Therefore, the magnitude of the force pressing the lawn mower directly into the ground is 25N.

b) To calculate the magnitude of the effective force that moves the lawn mower forward, we need to find the horizontal component of the force. The horizontal component is responsible for the forward motion of the mower.

By using trigonometry, we can find the horizontal component:

Horizontal Component = Force × cos(angle)
Horizontal Component = 50N × cos(30°)
Horizontal Component = 50N × √3/2
Horizontal Component ≈ 43.30N

Therefore, the magnitude of the effective force that moves the lawn mower forward is approximately 43.30N.

c) The lawn mower moves forward and not downward into the ground because of Newton's first law of motion, also known as the law of inertia. According to this law, an object at rest or in motion will remain in its current state unless acted upon by an external force.

In this case, the force being exerted on the lawn mower is at an angle, resulting in a vertical and horizontal component. The vertical component presses the lawn mower into the ground, but since the angle is designed to propel the mower forward rather than downward, the horizontal component contributes to the mower's forward motion.

Additionally, the horizontal component of the force overcoming friction between the mower's wheels and the ground helps propel the lawn mower forward. This force overcomes the resistance of friction, allowing the mower to move along the ground.