Mechanical Engineering: Ch 11: Friction (4 of 47) Forces on an Inclined Plane: Ex. 1 By Michel van Biezen
Summary by www.lecturesummary.com: Mechanical Engineering: Ch 11: Friction (4 of 47) Forces on an Inclined Plane: Ex. 1 By Michel van Biezen
Inclined Plane Example
(00:00 - 00:15)
- Force acting on an object sitting on an inclined plane
- Component of force pushing object up incline equals component of weight pushing object down incline
Balanced Forces
(00:11 - 00:25)
- Component of force pushing object up incline equals component of weight pushing object down incline
- This means there are no friction forces acting between the object and the surface
Friction Forces
(00:23 - 00:37)
- Maximum friction force = normal force x coefficient of static friction
- Since the two components are equal, there is no net force trying to push the object up or down the incline
- Therefore, the friction force is zero
Forces on the Object
(00:35 - 00:48)
- Weight of the object = 200 Newtons pushing straight down
- Force pushing object horizontally from right to left
- Reaction force perpendicular to the inclined plane
Reaction Force
(00:46 - 01:00)
- Reaction force is equal to the normal force
- Reaction force = mg cosine(θ) + F sin(θ)
- Angle θ is the same for all angles in the diagram
Solving for Force and Reaction Force
(00:57 - 01:09)
- Can use traditional method of finding parallel and perpendicular components
- Or can use trigonometric relationships (cosine and tangent) to solve for force and reaction force
Traditional Method
(01:07 - 01:22)
- F cos(θ) = mg sin(θ)
- F = (mg sin(θ)) / cos(θ) = mg tan(θ)
- Reaction force = mg cos(θ) + F sin(θ)
Trigonometric Method
(01:19 - 01:32)
- cos(θ) = adjacent/hypotenuse, so R = 100/cos(25°)
- tan(θ) = opposite/adjacent, so F = 100 tan(25°)
Comparison of Methods
(01:29 - 01:45)
- Both traditional and trigonometric methods give the same results
- Trigonometric method may be easier in some cases
Coefficients of Friction
(01:42 - 01:56)
- Coefficient of static friction is 0.35
- Coefficient of kinetic friction is 0.25
- Not needed for this example since the object is not moving
Solving for Force and Reaction Force
(01:54 - 02:07)
- Goal is to find the magnitude of the force and the reaction force
- Can use either the traditional or trigonometric method
Combining Forces
(02:04 - 02:17)
- The applied force, weight of the object, and reaction force should add up to zero
- The angle θ is the same for all angles in the diagram
Solving Using Trigonometry
(02:15 - 02:26)
- cos(θ) = adjacent/hypotenuse, so R = 100/cos(25°)
- tan(θ) = opposite/adjacent, so F = 100 tan(25°)
Solving Using Traditional Method
(02:25 - 02:39)
- F cos(θ) = mg sin(θ)
- F = (mg sin(θ)) / cos(θ) = mg tan(θ)
- Reaction force = mg cos(θ) + F sin(θ)
Comparing Methods
(02:36 - 02:50)
- Both the traditional and trigonometric methods give the same results
- The trigonometric method may be simpler in some cases
Calculating the Force
(02:47 - 03:02)
- F = mg tan(θ)
- F = 100 Newtons * tan(25°) = 46.6 Newtons
Calculating the Reaction Force
(02:58 - 03:12)
- Reaction force = mg cos(θ) + F sin(θ)
- Reaction force = 100 Newtons * cos(25°) + 46.6 Newtons * sin(25°) = 110.3 Newtons
Trigonometric Relationships
(03:08 - 03:24)
- cos(θ) = adjacent/hypotenuse
- tan(θ) = opposite/adjacent
Calculating the Force and Reaction Force
(03:19 - 03:34)
- F = 100 Newtons * tan(25°) = 46.6 Newtons
- Reaction force = 100 Newtons * cos(25°) + 46.6 Newtons * sin(25°) = 110.3 Newtons
Comparison of Methods
(03:29 - 03:43)
- Both the traditional and trigonometric methods give the same results for the force and reaction force
- The trigonometric method may be simpler in some cases
Reaction Force Calculation
(03:40 - 03:56)
- Reaction force = mg cos(θ) + F sin(θ)
- Plugging in the values: 100 Newtons * cos(25°) + 46.6 Newtons * sin(25°) = 110.3 Newtons
Conclusion
(03:50 - 04:05)
- The force and reaction force can be solved using either the traditional method or the trigonometric method
- The trigonometric method may be simpler in some cases
- The key is understanding the relationships between the forces and the angles involved
(00:00 - 00:15)
- Force acting on an object sitting on an inclined plane
- Component of force pushing object up incline equals component of weight pushing object down incline
Balanced Forces
(00:11 - 00:25)
- Component of force pushing object up incline equals component of weight pushing object down incline
- This means there are no friction forces acting between the object and the surface
Friction Forces
(00:23 - 00:37)
- Maximum friction force = normal force x coefficient of static friction
- Since the two components are equal, there is no net force trying to push the object up or down the incline
- Therefore, the friction force is zero
Forces on the Object
(00:35 - 00:48)
- Weight of the object = 200 Newtons pushing straight down
- Force pushing object horizontally from right to left
- Reaction force perpendicular to the inclined plane
Reaction Force
(00:46 - 01:00)
- Reaction force is equal to the normal force
- Reaction force = mg cosine(θ) + F sin(θ)
- Angle θ is the same for all angles in the diagram
Solving for Force and Reaction Force
(00:57 - 01:09)
- Can use traditional method of finding parallel and perpendicular components
- Or can use trigonometric relationships (cosine and tangent) to solve for force and reaction force
Traditional Method
(01:07 - 01:22)
- F cos(θ) = mg sin(θ)
- F = (mg sin(θ)) / cos(θ) = mg tan(θ)
- Reaction force = mg cos(θ) + F sin(θ)
Trigonometric Method
(01:19 - 01:32)
- cos(θ) = adjacent/hypotenuse, so R = 100/cos(25°)
- tan(θ) = opposite/adjacent, so F = 100 tan(25°)
Comparison of Methods
(01:29 - 01:45)
- Both traditional and trigonometric methods give the same results
- Trigonometric method may be easier in some cases
Coefficients of Friction
(01:42 - 01:56)
- Coefficient of static friction is 0.35
- Coefficient of kinetic friction is 0.25
- Not needed for this example since the object is not moving
Solving for Force and Reaction Force
(01:54 - 02:07)
- Goal is to find the magnitude of the force and the reaction force
- Can use either the traditional or trigonometric method
Combining Forces
(02:04 - 02:17)
- The applied force, weight of the object, and reaction force should add up to zero
- The angle θ is the same for all angles in the diagram
Solving Using Trigonometry
(02:15 - 02:26)
- cos(θ) = adjacent/hypotenuse, so R = 100/cos(25°)
- tan(θ) = opposite/adjacent, so F = 100 tan(25°)
Solving Using Traditional Method
(02:25 - 02:39)
- F cos(θ) = mg sin(θ)
- F = (mg sin(θ)) / cos(θ) = mg tan(θ)
- Reaction force = mg cos(θ) + F sin(θ)
Comparing Methods
(02:36 - 02:50)
- Both the traditional and trigonometric methods give the same results
- The trigonometric method may be simpler in some cases
Calculating the Force
(02:47 - 03:02)
- F = mg tan(θ)
- F = 100 Newtons * tan(25°) = 46.6 Newtons
Calculating the Reaction Force
(02:58 - 03:12)
- Reaction force = mg cos(θ) + F sin(θ)
- Reaction force = 100 Newtons * cos(25°) + 46.6 Newtons * sin(25°) = 110.3 Newtons
Trigonometric Relationships
(03:08 - 03:24)
- cos(θ) = adjacent/hypotenuse
- tan(θ) = opposite/adjacent
Calculating the Force and Reaction Force
(03:19 - 03:34)
- F = 100 Newtons * tan(25°) = 46.6 Newtons
- Reaction force = 100 Newtons * cos(25°) + 46.6 Newtons * sin(25°) = 110.3 Newtons
Comparison of Methods
(03:29 - 03:43)
- Both the traditional and trigonometric methods give the same results for the force and reaction force
- The trigonometric method may be simpler in some cases
Reaction Force Calculation
(03:40 - 03:56)
- Reaction force = mg cos(θ) + F sin(θ)
- Plugging in the values: 100 Newtons * cos(25°) + 46.6 Newtons * sin(25°) = 110.3 Newtons
Conclusion
(03:50 - 04:05)
- The force and reaction force can be solved using either the traditional method or the trigonometric method
- The trigonometric method may be simpler in some cases
- The key is understanding the relationships between the forces and the angles involved