equations of equilibrium, all work, and units 24 points): Determine the magnitude of force F and P. 1oolb y •FBD: F) angle cales? 100 lb QO (-3) 110° Fay 55 Ca15) 3 ఆర్భ) 4 105P F 55° You dont knon Fand P..? Pardllelogram neorrect t not needed here Calculations: Fx = Ecos Ca15°)= -54.72 lbs ? Fy -F sin cai5°)= -38,31|bs P PC415)=-89.59/bs Py3DPC315)%3D-67.2 lbs anafropridte Sin Clo5°) Sin Ca150) 66.Slbs 64.231bs= P.sin Ca15°) P= +l1.98 or ll2 ibs in Ibs Fat Fa 100 Cos CI00)= -34.20lbsv -y directi- on Fay lo0lbs sin Ca15°) = -57.35 |bs Fax+ - 가 incorrect. show E of E C-34.20 1bs) + C-57.35|bs)- F
equations of equilibrium, all work, and units 24 points): Determine the magnitude of force F and P. 1oolb y •FBD: F) angle cales? 100 lb QO (-3) 110° Fay 55 Ca15) 3 ఆర్భ) 4 105P F 55° You dont knon Fand P..? Pardllelogram neorrect t not needed here Calculations: Fx = Ecos Ca15°)= -54.72 lbs ? Fy -F sin cai5°)= -38,31|bs P PC415)=-89.59/bs Py3DPC315)%3D-67.2 lbs anafropridte Sin Clo5°) Sin Ca150) 66.Slbs 64.231bs= P.sin Ca15°) P= +l1.98 or ll2 ibs in Ibs Fat Fa 100 Cos CI00)= -34.20lbsv -y directi- on Fay lo0lbs sin Ca15°) = -57.35 |bs Fax+ - 가 incorrect. show E of E C-34.20 1bs) + C-57.35|bs)- F
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
I want to see what I did wrong on this promblem and straighen this out
![**Problem Statement:**
Determine the magnitude of forces \( F \) and \( P \).
**Free Body Diagram (FBD):**
- A diagram displays a vector at 100 lb at an angle of 110° in the X-Y plane.
- Forces are resolved into their X and Y components, labeled as \( F_{ax} \) and \( F_{ay} \), respectively.
**Diagram Explanation:**
- The FBD shows vectors in a coordinate system with angles indicated.
- A triangulated setup with angle markings for vectors is used to resolve the forces.
- There are notes indicating corrections on the diagram, such as incorrect parallelogram assumptions and inappropriate angle calculations.
**Calculations:**
1. **Resolving Forces:**
\[
F_x = F \cos(215^\circ) = -54.72 \text{ lbs}
\]
\[
F_y = F \sin(215^\circ) = -38.31 \text{ lbs}
\]
2. **Components of \( P \):**
\[
P_x = P \cos(45^\circ) = -89.59 \text{ lbs}
\]
\[
P_y = P \cos(315^\circ) = -67.2 \text{ lbs}
\]
3. **Components of the 100 lb Force:**
\[
F_{ax} = 100 \cos(110^\circ) = -34.20 \text{ lbs}
\]
\[
F_{ay} = 100 \sin(215^\circ) = -57.35 \text{ lbs}
\]
4. **Resultant Force Calculation:**
\[
\sqrt{(-34.20)^2 + (-57.35)^2} = F
\]
\[
F = 66.81 \text{ lbs in the negative X and Y direction}
\]
5. **Resolving \( P \) using Sine Law:**
\[
\frac{\sin(105^\circ)}{P} = \frac{\sin(215^\circ)}{66.81}
\]
\[
P = 111.98 \text{ or } 112 \text{ lbs in the negative Y direction}
\]
**Notes and Corrections:**
- Incorrect process indicated;](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F308db737-7862-446c-973d-78bf4e1d9584%2Fc9cf8781-8e3e-460c-b2a2-d8e21a123c24%2F1f7bseg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Determine the magnitude of forces \( F \) and \( P \).
**Free Body Diagram (FBD):**
- A diagram displays a vector at 100 lb at an angle of 110° in the X-Y plane.
- Forces are resolved into their X and Y components, labeled as \( F_{ax} \) and \( F_{ay} \), respectively.
**Diagram Explanation:**
- The FBD shows vectors in a coordinate system with angles indicated.
- A triangulated setup with angle markings for vectors is used to resolve the forces.
- There are notes indicating corrections on the diagram, such as incorrect parallelogram assumptions and inappropriate angle calculations.
**Calculations:**
1. **Resolving Forces:**
\[
F_x = F \cos(215^\circ) = -54.72 \text{ lbs}
\]
\[
F_y = F \sin(215^\circ) = -38.31 \text{ lbs}
\]
2. **Components of \( P \):**
\[
P_x = P \cos(45^\circ) = -89.59 \text{ lbs}
\]
\[
P_y = P \cos(315^\circ) = -67.2 \text{ lbs}
\]
3. **Components of the 100 lb Force:**
\[
F_{ax} = 100 \cos(110^\circ) = -34.20 \text{ lbs}
\]
\[
F_{ay} = 100 \sin(215^\circ) = -57.35 \text{ lbs}
\]
4. **Resultant Force Calculation:**
\[
\sqrt{(-34.20)^2 + (-57.35)^2} = F
\]
\[
F = 66.81 \text{ lbs in the negative X and Y direction}
\]
5. **Resolving \( P \) using Sine Law:**
\[
\frac{\sin(105^\circ)}{P} = \frac{\sin(215^\circ)}{66.81}
\]
\[
P = 111.98 \text{ or } 112 \text{ lbs in the negative Y direction}
\]
**Notes and Corrections:**
- Incorrect process indicated;
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