3. FBD for AK to determine the internal forces on K: forces/reactions on the AK; a. There are b. Sum axial forces to zero, solve that nominal force FK = c. Sum tangential forces to zero, solve that shear force VK d. Sum moment about point = N; direction is pointing to the N; direction is pointing to the solve that bending moment on k is MK = Nm; direction is pointing to the
Plane Trusses
It is defined as, two or more elements like beams or any two or more force members, which when assembled together, behaves like a complete structure or as a single structure. They generally consist of two force member which means any component structure where the force is applied only at two points. The point of contact of joints of truss are known as nodes. They are generally made up of triangular patterns. Nodes are the points where all the external forces and the reactionary forces due to them act and shows whether the force is tensile or compressive. There are various characteristics of trusses and are characterized as Simple truss, planar truss or the Space Frame truss.
Equilibrium Equations
If a body is said to be at rest or moving with a uniform velocity, the body is in equilibrium condition. This means that all the forces are balanced in the body. It can be understood with the help of Newton's first law of motion which states that the resultant force on a system is null, where the system remains to be at rest or moves at uniform motion. It is when the rate of the forward reaction is equal to the rate of the backward reaction.
Force Systems
When a body comes in interaction with other bodies, they exert various forces on each other. Any system is under the influence of some kind of force. For example, laptop kept on table exerts force on the table and table exerts equal force on it, hence the system is in balance or equilibrium. When two or more materials interact then more than one force act at a time, hence it is called as force systems.
![## Problem Statement:
Knowing that the radius of each pulley is 200 mm and neglecting friction, an external force is 360 N. Determine the internal forces at Points J and K of the frame shown.
### Solution:
#### 1. FBD for Frame and Pulleys Together:
- **a. Support Reactions:**
- There are [ ] supports on the frame, they are both [ ]; all together there are [ ] reactions on the supports. For a 2D body, this is statically [ ].
- **b. Moment Calculation:**
- Sum moment about point [ ], solve that \( B_x = \) [ ] N; direction is pointing to the [ ].
- **c. Horizontal Forces:**
- Sum all the horizontal forces to zero, solve that \( A_x = \) [ ] N; direction is pointing to the [ ].
- **d. Vertical Forces:**
- Sum all the vertical forces to zero, yield that \( A_y + B_y = \) [ ] N.
#### 2. FBD for Member AE:
- **a. Reaction Forces:**
- There are [ ] forces/reactions on the member AE. For a 2D body, this is statically [ ].
- **b. Moment Calculation:**
- Sum moment about point [ ], and note that the tension force in the cable on D is [ ]; solve that \( A_y = \) [ ].
- **c. Vertical Reaction:**
- Thus \( B_y = \) [ ] N; direction is pointing to the [ ].
- **d. Horizontal Forces:**
- Sum all the horizontal forces to zero, solve that \( E_x = \) [ ] N; direction is pointing to the [ ].
#### 3. FBD for AK to Determine the Internal Forces on K:
- **a. Internal Forces:**
- There are [ ] forces/reactions on the AK.
### Diagram Explanation:
The diagram on the right shows a frame with two pulleys and several key points labeled A, B, C, D, E, J, and K. A force is applied downward on point E, with each pulley having a specified radius. The distances between the points are also labeled, providing the necessary measurements to analyze the structure and solve for internal forces.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a42e238-4116-4e80-81a7-f4d567b4bd0e%2F536fe76a-c9f6-4c80-9524-19da2c30d07e%2Fb8r7s8_processed.jpeg&w=3840&q=75)
![**Transcription of Educational Content:**
### 3. FBD for AK to Determine the Internal Forces on K:
a. There are [ ] forces/reactions on the AK;
b. Sum axial forces to zero, solve that nominal force \( F_K = [ ] \) N; direction is pointing to the [ ].
c. Sum tangential forces to zero, solve that shear force \( V_K = [ ] \) N; direction is pointing to the [ ].
d. Sum moment about point [ ], solve that bending moment on K is \( M_K = [ ] \) Nm; direction is pointing to the [ ].
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### 4. FBD for BJ to Determine the Internal Forces on J:
a. There are [ ] forces/reactions on the AK;
b. Sum axial forces to zero, solve that nominal force \( F_K = [ ] \) N; direction is pointing to the [ ].
c. Sum tangential forces to zero, solve that shear force \( V_K = [ ] \) N; direction is pointing to the [ ].
d. Sum moment about point [ ], solve that bending moment on J is \( M_J = [ ] \) Nm; direction is pointing to the [ ].
### Diagram/Graph Explanation:
The form involves free body diagrams (FBDs) for sections AK and BJ, used to compute internal forces and moments. It asks for the completion of parts related to axial, tangential forces and bending moments, with blanks for input. Drop-down menus likely allow for selecting directions of forces.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a42e238-4116-4e80-81a7-f4d567b4bd0e%2F536fe76a-c9f6-4c80-9524-19da2c30d07e%2F0qebehd_processed.jpeg&w=3840&q=75)
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