A wheelbarrow inclined at 30° to the horizontal is pushed with a force of 150N. (A) what is the horizontal and vertical components of the force. (B) in what direction will the wheelbarrow move and why.
To find the horizontal and vertical components of the force, we can use trigonometry.
A)
Given:
Angle of inclination (θ) = 30°
Force applied (F) = 150N
The horizontal component of the force (F_horizontal) is given by F * cos(θ):
F_horizontal = F * cos(θ)
F_horizontal = 150N * cos(30°)
F_horizontal = 150N * 0.866
F_horizontal = 129.9N
The vertical component of the force (F_vertical) is given by F * sin(θ):
F_vertical = F * sin(θ)
F_vertical = 150N * sin(30°)
F_vertical = 150N * 0.5
F_vertical = 75N
So, the horizontal component of the force is approximately 129.9N, and the vertical component is approximately 75N.
B)
To determine the direction in which the wheelbarrow will move, we need to consider the forces acting on it. Since the force applied has both horizontal and vertical components, it will contribute to both the horizontal and vertical motion of the wheelbarrow.
In this case, the horizontal component of the force is greater than the vertical component. Therefore, the wheelbarrow will primarily move horizontally in the direction of the applied force. However, it will also experience a slight upward motion due to the vertical component of the force.
So, the wheelbarrow will move primarily in the direction of the applied force (horizontally) with a slight upward motion due to the inclined angle.