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`        The force f1 parallel to incline plane that is necessary to move a body up the inclined is double the force f2 to that is necessary to just prevent them sliding down them`
one year ago

```							Let’s consider the 2nd case. F2 is the force required just to prevent the body from sliding. Now, if we consider the FBD, there will be F2 the direction opposite to wsinθ(component of weight along the incline). The friction force will act in the direction opposite to wsinθ, as the body tends to move down the incline, and F2 prevents it.Thus, we get F2=wsinθ-fr (fr = frictional force)———— 1As ? is the angle of repose, μ = tan?....2From 1, F2=wsinθ - μwcosθ (fr = μN, and N is the normal reaction, component for weight perpendicular to the incline)So, F2=wsinθ – tan?cosθPut tan?=sin?/cos?Thus,F2=wsinθ – (sin?cosθ)/cos?On further solving, we get F2 = wsin(θ - ?)sec?, that is option ‘A’ which is the answer. Hope this helped you. Good Luck.
```
one year ago
```							Let’s consider the 2nd case. F2 is the force required just to prevent the body from sliding. Now, if we consider the FBD, there will be F2 the direction opposite to wsinθ(component of weight along the incline). The friction force will act in the direction opposite to wsinθ, as the body tends to move down the incline, and F2 prevents it.Thus, we get F2=wsinθ-fr (fr = frictional force)———— 1As ? is the angle of repose, μ = tan?....2From 1, F2=wsinθ - μwcosθ (fr = μN, and N is the normal reaction, component for weight perpendicular to the incline)So, F2 = wsinθ – wtan?cosθPut tan? = sin?/cos?Thus,F2 = wsinθ – w(sin?cosθ)/cos?On further solving, we get F2 = wsin(θ - ?)sec?, that is option ‘A’ which is the answer. Hope this helped you. Good Luck.
```
one year ago
```							Sorry, but the first solution had some minute errors, so I reanswered... The second answer is correct.
```
one year ago
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