a crate of cooldrinks with a mass of 15kg over a smooth horizontal surface by exerting a force of 80N at 30° to the horizontal.
Draw a diagram and aswers the following
1. Draw a force diagram showing and labelling all the forces acting on the crate.
2. Calculate the vertical component of the force.
3. Calculate the force with which the crate is pulled across the surface.
2) 80sin30
3) Um. You said an 80 N force...
Sure! I can help you with that. Let's go through each question one by one.
1. To draw a force diagram, we need to identify all the forces acting on the crate. In this case, we have:
- Weight (mg): The force exerted by gravity on the crate. It acts vertically downward and can be calculated by multiplying the mass of the crate (15 kg) by gravity (9.8 m/s^2).
- Applied force (F): The force exerted by pushing the crate. It acts at an angle of 30° to the horizontal.
- Normal force (N): The force exerted by the surface on the crate perpendicular to the surface. Since the crate is on a smooth horizontal surface and moving horizontally, the normal force cancels out the weight of the crate.
2. To calculate the vertical component of the force (Fv), we need to find the projection of the applied force in the vertical direction. We can use the formula Fv = F * sin(theta), where theta is the angle (30°) and F is the magnitude of the force (80 N). Plug in the values and calculate:
Fv = 80 N * sin(30°)
Fv ≈ 40 N
Therefore, the vertical component of the force is approximately 40 N.
3. To calculate the force with which the crate is pulled across the surface, we need to find the horizontal component of the applied force (Fh). We can use the formula Fh = F * cos(theta), where theta is the angle (30°) and F is the magnitude of the force (80 N). Plug in the values and calculate:
Fh = 80 N * cos(30°)
Fh ≈ 69.28 N
Therefore, the force with which the crate is pulled across the surface is approximately 69.28 N.