A warehouse worker pushes a crate along the floor, as shown in the figure , by a force of 10.0 N that points downward at an angle of 45.0 degree below the horizontal.
A. Find the horizontal component of the push.
B.Find the vertical component of the push.
7.07n
To find the horizontal and vertical components of the push, we need to break down the given force into its horizontal and vertical components using trigonometry.
Let's denote the force pushing the crate as F = 10.0 N and the angle it makes with the horizontal as θ = 45.0 degrees below the horizontal.
A. To find the horizontal component of the push, we can use the cosine function. The horizontal component (Fx) can be calculated as:
Fx = F * cos(θ)
Plugging in the values, we have:
Fx = 10.0 N * cos(45.0 degrees)
= 10.0 N * 0.7071 (rounded to four decimal places)
Calculating this, we get:
Fx ≈ 7.071 N (rounded to three decimal places)
So, the horizontal component of the push is approximately 7.071 N.
B. To find the vertical component of the push, we can use the sine function. The vertical component (Fy) can be calculated as:
Fy = F * sin(θ)
Plugging in the values, we have:
Fy = 10.0 N * sin(45.0 degrees)
= 10.0 N * 0.7071 (rounded to four decimal places)
Calculating this, we get:
Fy ≈ 7.071 N (rounded to three decimal places)
So, the vertical component of the push is approximately 7.071 N.
Therefore, the horizontal component of the push is 7.071 N and the vertical component of the push is 7.071 N.