Question:

A 10 kg box is resting on a flat surface. A force is applied to the box at an angle of 30 degrees above the horizontal. The force has a magnitude of 50 N. Draw a free-body diagram of the box and calculate the net force and its components on the box.

Answer:

The free-body diagram shows the forces acting on the box:

• Weight (W): A downward force due to gravity with a magnitude of 98 N (10 kg multiplied by the acceleration due to gravity, g = 9.8 m/s²).
• Normal force (N): An upward force exerted by the surface to support the weight of the box. It has the same magnitude as the weight, but in the opposite direction.
• Applied force (F): A force of 50 N acting at a 30-degree angle above the horizontal.

To calculate the net force and its components, we need to resolve the forces into their horizontal and vertical components.

Let's consider the components of the applied force:

• F_x = F × cos(θ) = 50 N × cos(30°) = 43.3 N (to the right)
• F_y = F × sin(θ) = 50 N × sin(30°) = 25 N (upward)

Next, let's consider the weight and the normal force:

• Weight (W) is only acting in the vertical direction, so W_x = 0 N and W_y = -98 N (downward).
• The normal force (N) should counterbalance the weight, so N_x = 0 N and N_y = 98 N (upward).

Now, let's find the net force components:

• F_net,x = F_x + N_x = 43.3 N + 0 N = 43.3 N (to the right)
• F_net,y = F_y + N_y + W_y = 25 N + 98 N - 98 N = 25 N (upward)

Therefore, the net force acting on the box is 43.3 N to the right and 25 N upward.