Let's solve problem **f** step-by: --- ## Problem Restatement **f.)** If possible, determine the value of \( F_E \) such that the door remains in static equilibrium. Also draw the \( \mathbf{r} \) vector for \( F_E \) (the moment arm). - The door is 1 m long. - A 100 N force acts **upwards** at the center (at .5 m from hinge A). - \( F_E \) acts at the end **E** (1 m from A), at a \( 45^\circ \) angle **up and right**. - We need to find the value of \( F_E \) such that the sum of moments about point A is zero (static equilibrium). --- ## Step 1: Draw Forces and Moment Arms - **100 N Force**: Acts at B (.5 m from A), vertically upward. - **\( F_E \) Force**: Acts at E (1 m from A), at \( 45^\circ \) up and right. --- ## Step 2: Write Moment Equations Take moments about point \( A \): \[ \sum M_A = \] ### Contribution of 100 N Force - Perpendicular distance from A: \( r_{100} = .5 \) m - Force: 100 N (upward) - Moment: \( +100 \times .5 = +50 \) Nm (counterclockwise is positive) ### Contribution of \( F_E \) - Position vector from A to E: \( \mathbf{r}_E = 1 \) m - \( F_E \) acts at \( 45^\circ \) above the door. The perpendicular component of \( F_E \) to the door is: \[ F_{E,\perp} = F_E \sin(45^\circ) = F_E \frac{\sqrt{2}}{2} \] The moment arm is 1 m. The *direction* of the moment: Since \( F_E \) is "up and right" at E, it tends to rotate the door **clockwise** about A, so negative. \[ \text{Moment from } F_E = - F_E \sin(45^\circ) \times 1 \] --- ## Step 3: Set Up the Equation \[ \sum M_A = = +50 - F_E \sin(45^\circ) \times 1 \] \[ F_E \sin(45^\circ) = 50 \] --- ## Step 4: Solve for \( F_E \) \[ F_E = \frac{50}{\sin(45^\circ)} \] \[ \sin(45^\circ) = \frac{\sqrt{2}}{2} \approx .7071 \] \[ F_E = \frac{50}{.7071} \approx 70.7 \text{ N} \] --- ## Step 5: Check Direction - The diagram shows the answer as **negative** (\(-70.7\) N). - This indicates that to balance the moment, \( F_E \) must act **down and to the left** (opposite to the drawn arrow). - The sign convention is important: the way we've set it up, a positive \( F_E \) is "up and right" (as shown), but the moment it produces is clockwise (negative). To balance the 100 N force's counterclockwise moment, \( F_E \) must act "down and left" (opposite the arrow shown), i.e., \( F_E = -70.7 \) N. --- ## **Final Answer** \[ \boxed{F_E = -70.7 \text{ N}} \] - The negative sign means the direction is **down and to the left**, opposite to the arrow shown in the diagram. - The \( \mathbf{r} \) vector is from point \( A \) (hinge) to point \( E \) (end of the door), i.e., a 1 m vector along the door. --- ### **Summary Table** | Force | Location | Distance from A (m) | Direction | Moment About A (Nm) | |------------|----------|---------------------|--------------------|----------------------| | 100 N | B | .5 | Up | +50 (counterclockwise) | | \( F_E \) | E | 1. | Down-left (\(-70.7\) N) | -50 (clockwise) | --- ### **Free Body Diagram Explanation** - \( \mathbf{r} \) for \( F_E \) is a vector from A to E (1 m along the door). - \( F_E \) must be applied down and left at E (opposite to the red arrow in the diagram), with a magnitude of 70.7 N, to keep the door in static equilibrium. --- If you need the diagrams redrawn or more explanation, let me know!

# Problem Overview We need to analyze the static equilibrium of a door subjected to external forces. Specifically, we will determine the force \( F_E \) that maintains the door in static equilibrium under the influence of a 100 N upward force applied at the center of the door. ## Given Data - Length of the door: \( L = 1 \) m - Force \( F \) at midpoint (0.5 m from hinge): \( 100 \) N (upward) - Force \( F_E \) at the end (1 m from hinge): Applied at \( 45^\circ \) (up and right) ## Step 1: Identify Forces and Moment Arms 1. **100 N Force**: - Location: Midpoint (0.5 m from hinge A) - Direction: Upward 2. **Force \( F_E \)**: - Location: End of the door (1 m from hinge A) - Angle: \( 45^\circ \) up and to the right ## Step 2: Moment Calculation About Point A To maintain static equilibrium, the sum of moments about point A must equal zero: \[ \sum M_A = 0 \] ### Contribution of the 100 N Force - Moment arm: \( r_{100} = 0.5 \) m (perpendicular distance) - Moment due to 100 N force (counterclockwise is positive): \[ M_{100} = +100 \times 0.5 = +50 \, \text{Nm} \] ### Contribution of Force \( F_E \) - Perpendicular component of \( F_E \): \[ F_{E,\perp} = F_E \sin(45^\circ) = F_E \frac{\sqrt{2}}{2} \] - Moment arm: \( r_E = 1 \) m - Moment due to \( F_E \) (clockwise is negative): \[ M_E = - F_E \sin(45^\circ) \times 1 \] ## Step 3: Set Up the Equation Setting the net moment to zero: \[ \sum M_A = 0 \Rightarrow 50 - F_E \sin(45^\circ) = 0 \] This leads to: \[ F_E \sin(45^\circ) = 50 \] ## Step 4: Solve for \( F_E \) Substituting \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \): \[ F_E \frac{\sqrt{2}}{2} = 50 \] Rearranging gives: \[ F_E = \frac{50 \times 2}{\sqrt{2}} = \frac{100}{\sqrt{2}} \approx 70.71 \, \text{N} \] ## Step 5: Direction Verification The positive value indicates that \( F_E \) is acting **up and to the right**. However, to balance the moment from the 100 N force, \( F_E \) must act **down and to the left**: \[ F_E = -70.71 \, \text{N} \] ## Final Answer \[ \boxed{F_E = -70.71 \, \text{N}} \] - The negative sign indicates the direction is **down and to the left**. --- ## Summary of Forces | Force | Location | Distance from A (m) | Direction | Moment About A (Nm) | |------------|----------|---------------------|-------------------------|--------------------------| | 100 N | B | 0.5 | Up | +50 (counterclockwise) | | \( F_E \) | E | 1 | Down and to the left | -50 (clockwise) | --- ### Moment Arm Vector - The position vector \( \mathbf{r} \) for \( F_E \) is from point A to E, representing a 1 m vector along the length of the door. ### Free Body Diagram - A diagram should be drawn showing the forces acting on the door, with the \( \mathbf{r} \) vector pointing from the hinge to the end of the door and indicating the direction of \( F_E \). If additional clarification or diagrams are required, please let me know!