SOLUTION: A body of weight 500N rests on a plane inclined at 20o to the horizontal. The coefficient of friction
is 0.4, determine a force F at an angle of 15o to the plane required to
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Question 1209832: A body of weight 500N rests on a plane inclined at 20o to the horizontal. The coefficient of friction
is 0.4, determine a force F at an angle of 15o to the plane required to
(a) Pull the body upwards
(b) Push the body downwards
(c) Pull the body downwards
(d) Push the body upwards
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down this problem step-by-step.
**1. Define the Forces**
* **Weight (W):** 500 N (acting vertically downwards)
* **Normal Force (N):** Perpendicular to the inclined plane
* **Friction Force (f):** Opposing the motion, f = μN (where μ = 0.4)
* **Applied Force (F):** At 15° to the plane
* **Angle of Inclination (θ):** 20°
**2. Resolve the Weight into Components**
* **Component parallel to the plane (W_parallel):** W sin(θ) = 500 sin(20°) ≈ 171.01 N
* **Component perpendicular to the plane (W_perpendicular):** W cos(θ) = 500 cos(20°) ≈ 469.85 N
**3. Resolve the Applied Force into Components**
* **Component parallel to the plane (F_parallel):** F cos(15°)
* **Component perpendicular to the plane (F_perpendicular):** F sin(15°)
**4. Calculate the Normal Force (N)**
* N = W_perpendicular + F_perpendicular = 469.85 + F sin(15°)
**5. Calculate the Friction Force (f)**
* f = μN = 0.4(469.85 + F sin(15°)) = 187.94 + 0.4F sin(15°)
**6. Solve for F in Each Scenario**
**(a) Pull the body upwards**
* The forces parallel to the plane must balance: F_parallel - W_parallel - f = 0
* F cos(15°) - 171.01 - (187.94 + 0.4F sin(15°)) = 0
* F cos(15°) - 171.01 - 187.94 - 0.4F sin(15°) = 0
* F(cos(15°) - 0.4 sin(15°)) = 358.95
* F(0.9659 - 0.1035) = 358.95
* F(0.8624) = 358.95
* F ≈ 416.22 N
**(b) Push the body downwards**
* The forces parallel to the plane must balance: F_parallel + W_parallel - f = 0
* F cos(15°) + 171.01 - (187.94 + 0.4F sin(15°)) = 0
* F(cos(15°) - 0.4 sin(15°)) = 16.93
* F(0.8624) = 16.93
* F ≈ 19.63 N
**(c) Pull the body downwards**
* The forces parallel to the plane must balance: W_parallel - F_parallel - f = 0
* 171.01 - F cos(15°) - (187.94 + 0.4F sin(15°)) = 0
* -F(cos(15°) + 0.4 sin(15°)) = 16.93
* F(cos(15°) + 0.4 sin(15°)) = -16.93
* F(0.9659 + 0.1035) = -16.93
* F(1.0694) = -16.93
* F ≈ -15.83 N
* Since F is negative, this is not possible without an external force preventing the object from sliding down. The force required to keep the body from accelerating downwards is 15.83 N.
**(d) Push the body upwards**
* The forces parallel to the plane must balance: W_parallel - F_parallel + f = 0
* 171.01 - F cos(15°) + (187.94 + 0.4F sin(15°)) = 0
* -F(cos(15°) - 0.4 sin(15°)) = -358.95
* F(cos(15°) - 0.4 sin(15°)) = 358.95
* F(0.8624) = 358.95
* F ≈ 416.22 N
**Summary**
* **(a) Pull upwards:** F ≈ 416.22 N
* **(b) Push downwards:** F ≈ 19.63 N
* **(c) Pull downwards:** F ≈ 15.83 N (to prevent downward acceleration)
* **(d) Push upwards:** F ≈ 416.22 N
Answer by ikleyn(52786) (Show Source): You can put this solution on YOUR website!
.
The solution to this problem in the post by @CPhill is INCORRECT.
The fact is that the applied force at 15° to the inclined plane DECREASES the normal force
produced by the weight.
In the solution by @CPhill the applied force at 15° to the inclined plane, in opposite,
INCREASES the normal force produced by the weight, which is physically absurdist.
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