Drawing Free Body Diagrams for Engineering Statics

This is a quick slide show I put together for a friend. I'll have more to come. I taught elementary and engineering physics (algebra based) at Texas State University in San Marcos. Most of the slide shows I plan to do will be physics related

1. ENGINEERING Statics

2. NET FORCE When the net force on an object is zero, the ACCELERATION of the object is zero, However, the object may still be moving IN A STRAIGHT LINE

3. FUNDAMENTAL FORCES How many forces are involved in a free body diagram? 1 2 3 4 …. ???

4. FUNDAMENTAL FORCES How many forces are involved in a free body diagram? 1 Gravity 2 Electromagnetic 3 Strong Nuclear Force 4 Weak Nuclear Force

5. FUNDAMENTAL FORCES How many forces are involved in a free body diagram? 1 Gravity 2 Electromagnetic 3 Strong Nuclear Force 4 Weak Nuclear Force The strong and weak nuclear forces have very short ranges and are only effective inside the nucleus. Therefore, you only have TWO fundamental forces to consider! Easy, right??

6. FUNDAMENTAL FORCES How many forces are involved in a free body diagram? 1 Gravity In most cases, one of the two masses is the earth (d is the radius of the earth, and Gm 1 /d 2 is “little g” or approximately 9.8 m/s 2 ). Even placing an object at the top of a tall skyscraper has little effect on gravity. The acceleration due to gravity is usually a known quantity given in the problem. 2 Electromagnetic  

7. FUNDAMENTAL FORCES How many forces are involved in a free body diagram? 1 Gravity 2 Electromagnetic Coulomb force Magnetic force Friction Tension Bouyant Normal Spring Intermolecular  

8. FUNDAMENTAL FORCES How many forces are involved in a free body diagram? 1 2 3 4

9. FUNDAMENTAL FORCES How many forces are involved in a free body diagram? Only one, possibly!  

10. WHY IS THIS SO HARD? Resolve all force vectors into their x y and z components Draw each pair of forces involved ONE PAIR AT A TIME   Let’s make it simpler

11. Gravity (assume the ropes have negligible weight. Then the force of gravity on them is zero. You may exclude these two pairs of forces) Intuitively, we might assume that the force through each rope is 50N. But we can also prove it mathematically. A simple substitution gives us the answer.  

12. Gravity (assume the ropes have negligible weight. Then the force of gravity on them is zero. You may exclude these two pairs of forces) According to Newton’s 3 rd Law of Gravity, forces always occur in pairs. What object does the “reaction force” act on? What is the magnitude and direction of this “reaction force”?

13. Gravity (assume the ropes have negligible weight. Then the force of gravity on them is zero. You may exclude these two pairs of forces) Since we are only concerned with the forces acting on the sign and not the earth, we can exclude the earth and the reaction force on it.

14. If the only force acting on the sign were the weight of the sign, there would be a NON-ZERO net force acting on the sign, and it would accelerate in the direction of the net force (in this case, towards the ground)

15. Tension Fortunately, there are 2 ropes supporting the weight of the sign. The SUM of these forces that are acting on the sign ADD UP TO ZERO. We know this because the sign is NOT ACCELERATING

16. CHECK YOUR WORK Did we include all relevant forces? Suggestion – always start with gravity So far, we have drawn the diagram. In the next step, we will calculate the horizontal and vertical components of the tension forces in each rope.

17.       AND

18. WHAT DO WE KNOW = cos We know this because the acceleration (and hence the net force) in the horizontal direction is zero . Don’t confuse these forces for action /reaction pairs of forces! Remember that the “reaction” forces are acting on THE SIGN. An easy way to remember this is to pretend the sign were made of a stretchy material.  

19. WHAT DO WE KNOW In this problem, it is ESSENTIAL to solve for T 2 before proceeding to the horizontal forces. Once you’ve done enough of these problems, you’ll see that each one is a puzzle to be solved. = cos