A ware house worker pushed a crate along the floor, as shown in fig 4.34, by a force of 10N that points downward at an angle of 45 degrees below the horizontal. Find the horizontal and vertical components of the push.
both horizontal and down are 10 * sqrt2 /2 Newtons
because
sin 45 = cos 45 = sqrt2 /2
To find the horizontal and vertical components of the push force, we need to use trigonometry.
Given:
Force applied, F = 10 N
Angle below the horizontal, θ = 45 degrees
To find the horizontal component (Fx), we'll use the cosine function:
Fx = F * cos(θ)
Substituting the given values:
Fx = 10 N * cos(45 degrees)
Calculating the horizontal component:
Fx = 10 N * 0.707 ≈ 7.07 N
To find the vertical component (Fy), we'll use the sine function:
Fy = F * sin(θ)
Substituting the given values:
Fy = 10 N * sin(45 degrees)
Calculating the vertical component:
Fy = 10 N * 0.707 ≈ 7.07 N
Therefore, the horizontal component of the push force is approximately 7.07 N, and the vertical component is approximately 7.07 N.
To find the horizontal and vertical components of the push, we need to decompose the force into its horizontal and vertical components.
The given force is 10N at an angle of 45 degrees below the horizontal. Let's call the horizontal component "F_x" and the vertical component "F_y".
To find the horizontal component (F_x), we can use trigonometry. The horizontal component can be determined using the cosine of the angle. The formula is:
F_x = F * cos(θ)
where F is the magnitude of the force and θ is the angle between the force and the horizontal direction.
In this case, F = 10N and θ = 45 degrees.
F_x = 10N * cos(45°)
Using the value of cos(45°) = √2 / 2, we can substitute it into the equation:
F_x = 10N * (√2 / 2)
F_x = 10N * (√2 / 2)
Simplifying the equation:
F_x = 10N * (√2 / 2)
F_x = 10N * √2 / 2
F_x = 5√2 N
So, the horizontal component of the push is 5√2 N.
To find the vertical component (F_y), we use the sine of the angle. The vertical component can be determined using the formula:
F_y = F * sin(θ)
where F is the magnitude of the force and θ is the angle between the force and the horizontal direction.
In this case, F = 10N and θ = 45 degrees.
F_y = 10N * sin(45°)
Using the value of sin(45°) = √2 / 2, we can substitute it into the equation:
F_y = 10N * (√2 / 2)
F_y = 10N * (√2 / 2)
Simplifying the equation:
F_y = 10N * (√2 / 2)
F_y = 10N * √2 / 2
F_y = 5√2 N
So, the vertical component of the push is 5√2 N.
Therefore, the horizontal component of the push is 5√2 N and the vertical component of the push is also 5√2 N.