Question:

A block of mass 2 kg is placed on a rough inclined plane with an angle of 30 degrees to the horizontal. The coefficient of friction between the block and the inclined plane is 0.3. A force F is applied to the block parallel to the incline, causing it to move upwards with a constant velocity. Draw the free-body diagram for the block and calculate the magnitude of the applied force F.

Assumptions:

• The inclined plane is frictionless except for the area in contact with the block.
• The acceleration is zero, resulting in a constant velocity.

Assume g = 9.8 m/s², ignore air resistance, and use the following formula for the magnitude of the force of friction:

Frictional force (f_f) = coefficient of friction (μ) * normal force (N)

Answer:

The free-body diagram for the block on the inclined plane can be represented as follows:

Here is the step-by-step explanation to solve the problem:

1. Identify the forces acting on the block. In this case, we have the following forces:

   • Weight (mg): acting vertically downwards due to gravity, with a magnitude of 2 kg * 9.8 m/s² = 19.6 N.
   • Normal force (N): acting perpendicular to the inclined plane, opposing the weight. It has the same magnitude as the component of weight normal to the plane, which is N = mg * cos(30°) = 19.6 N * cos(30°) = 16.99 N.
   • Applied force (F): acting parallel to the incline and causing motion in the upward direction.
   • Frictional force (f_f): opposing the direction of motion, acting parallel to the incline.

2. Determine the frictional force. Using the formula f_f = μN, where μ = 0.3, we can calculate the frictional force as f_f = 0.3 * 16.99 N = 5.097 N.

3. Draw the free-body diagram, including the magnitudes and direction of each force as shown in the diagram above.

4. Analyze the forces along the vertical and parallel directions:

   • Vertical direction: the forces are balanced as the block is not moving vertically. The weight (19.6 N) is balanced by the normal force (16.99 N).
   • Parallel direction: since the block is moving upwards with a constant velocity, the net force parallel to the incline must be zero. Therefore, the applied force F must have the same magnitude as the frictional force f_f. Thus, F = 5.097 N.

Hence, the magnitude of the applied force F required to move the block upwards with constant velocity is 5.097 N.