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**Problem**: A uniform bar of mass M and length l is propped against a very slick vertical wall as shown. The angle between the wall and the upper end of the bar is θ. The force of static friction between the upper end of the bar and the wall is negligible, but the bar remains at rest (in equilibrium). If we take the pivot at the point where the bar touches the floor, which expression below is (Στ), where x is along the floor and y is along the wall?
1. ℓ (Fwsinθ + Mg/2 cosθ) = 0
2. ℓ (Mg/2 cosθ − Fwsinθ) = 0
3. ℓ (Mg/2 − Fw) = 0
4. ℓ (−fssinθ − ncosθ) = 0
5. ℓ (Mg/2 + Fw) = 0
6. ℓ (Fwsinθ − Mg/2 cosθ) = 0
7. ℓ (−Mg/2 cosθ − Fwsinθ) = 0
8. ℓ (fssinθ − ncosθ) = 0
9. ℓ (Fwcosθ + Mg/2 sinθ) = 0
10. ℓ (Fwcosθ − Mg/2 sinθ) = 0

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###### Problem Details

A uniform bar of mass M and length l is propped against a very slick vertical wall as shown. The angle between the wall and the upper end of the bar is θ. The force of static friction between the upper end of the bar and the wall is negligible, but the bar remains at rest (in equilibrium). If we take the pivot at the point where the bar touches the floor, which expression below is (Στ), where x is along the floor and y is along the wall?

1. ℓ (F_{w}sinθ + Mg/2 cosθ) = 0

2. ℓ (Mg/2 cosθ − F_{w}sinθ) = 0

3. ℓ (Mg/2 − F_{w}) = 0

4. ℓ (−f_{s}sinθ − ncosθ) = 0

5. ℓ (Mg/2 + F_{w}) = 0

6. ℓ (F_{w}sinθ − Mg/2 cosθ) = 0

7. ℓ (−Mg/2 cosθ − F_{w}sinθ) = 0

8. ℓ (f_{s}sinθ − ncosθ) = 0

9. ℓ (F_{w}cosθ + Mg/2 sinθ) = 0

10. ℓ (F_{w}cosθ − Mg/2 sinθ) = 0

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Our tutors have indicated that to solve this problem you will need to apply the Equilibrium in 2D - Ladder Problems concept. You can view video lessons to learn Equilibrium in 2D - Ladder Problems Or if you need more Equilibrium in 2D - Ladder Problems practice, you can also practice Equilibrium in 2D - Ladder Problems practice problems .

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Based on our data, we think this problem is relevant for Professor Lai's class at TEXAS.