+448 Tackling Friction in Free body diagrams, would appreciate your generous help.
Problem: A block with a mass of 10 kg is placed on a rough inclined plane that makes a 30° angle with the horizontal. The coefficient of static friction between the block and the plane is 0.4, and the coefficient of kinetic friction is 0.3. A force (F) is applied parallel to the incline to push the block upward.
+46 Let's analyze the forces acting on the block and determine the conditions for it to move upward.
1. Draw a Free Body Diagram:
Weight (W): Acts vertically downwards. Its magnitude is W=mg=10 kg×9.8 m/s2=98 N.
Normal Force (N): Acts perpendicular to the inclined plane, pushing the block away from the surface.
Force Applied (F): Acts parallel to the inclined plane, pushing the block upward.
Friction Force (f): Acts parallel to the inclined plane, opposing the motion or the tendency of motion.
2. Resolve the Weight into Components:
We need to resolve the weight into components parallel and perpendicular to the inclined plane:
Component parallel to the incline (W∥): W∥=Wsin(30°)=98 N×sin(30°)=98 N×0.5=49 N. This component acts downwards along the incline.
Component perpendicular to the incline (W⊥): W⊥=Wcos(30°)=98 N×cos(30°)=98 N×23≈84.87 N. This component acts perpendicular to the incline and is balanced by the normal force.
3. Determine the Normal Force:
Since there is no acceleration perpendicular to the inclined plane, the normal force balances the perpendicular component of the weight:
N=W⊥≈84.87 N.
4. Determine the Maximum Static Friction Force:
The maximum static friction force (fs,max) is the maximum force that can oppose the initiation of motion. It is given by:
fs,max=μsN=0.4×84.87 N≈33.95 N.
This force acts downwards along the incline, opposing the upward force F.
5. Condition for the Block to Start Moving Upward:
For the block to start moving upward, the applied force F must overcome both the component of the weight acting down the incline and the maximum static friction force:
F>W∥+fs,max
F>49 N+33.95 N
F>82.95 N
Therefore, the minimum force required to start pushing the block upward is slightly greater than 82.95 N.
6. Determine the Kinetic Friction Force (if the block is moving):
If the applied force F is large enough to overcome static friction and the block starts moving upward, the friction force becomes kinetic friction (fk). It is given by:
fk=μkN=0.3×84.87 N≈25.46 N.
This force acts downwards along the incline, opposing the upward motion.
Summary of Forces and Conditions:
Component of weight down the incline: 49 N
Maximum static friction force (opposing upward motion): ≈33.95 N
Kinetic friction force (opposing upward motion once moving): ≈25.46 N
To answer specific questions about the motion, you would need to provide the magnitude of the applied force (F). For example:
If F < 49 N: The block will remain at rest, and the static friction force will act upward along the incline with a magnitude equal to F, balancing the component of the weight.
If 49 N ≤ F ≤ 82.95 N: The block will remain at rest, and the static friction force will act downward along the incline with a magnitude equal to W∥−F.
If F > 82.95 N: The block will accelerate upward. The net force acting on the block will be F−W∥−fk.
Let me know if you have a specific value for the force F or if you have a particular question about the situation (e.g., "What is the minimum force required to start moving the block?", "What is the acceleration of the block if F = 100 N?").
