a wheelbarrow inclined at 30° to the horizontal is pushed with a force of 150N . in what direction will the wheelbarrow move and why?
To determine the direction in which the wheelbarrow will move, we need to consider the components of the force applied and their respective directions.
In this case, we have a force of 150N applied at an angle of 30° to the horizontal. Let's break down this force into its horizontal and vertical components.
Horizontal Component:
The horizontal component is given by F_horizontal = F × cos(angle).
F_horizontal = 150N × cos(30°) = 150N × √(3)/2 = 150N × 0.866 ≈ 129.9N.
Vertical Component:
The vertical component is given by F_vertical = F × sin(angle).
F_vertical = 150N × sin(30°) = 150N × 0.5 = 75N.
Now, let's consider the forces acting on the wheelbarrow. Along the horizontal plane, there is the force of friction opposing the motion, and there is no external force acting vertically.
Since the horizontal component of the applied force (129.9N) is greater than the opposing force of friction, the wheelbarrow will experience a net force in the forward direction. Hence, the wheelbarrow will move in the direction of the applied force, which is angled at 30° to the horizontal.
Therefore, the wheelbarrow will move forward and slightly upward, following the direction of the applied force at an angle of 30° to the horizontal.
Am I missing something? The wheelbarrow has to move in the direction of the net force. Newtons law of motion.
Hmm
But this question talks about inclined and horizontal
My thinking is that maybe horizontal formula will work like I meant Vy cos theta
150cos30°
Can it work