**Unformatted text preview: **OREGON STATE UNIVERSITY College of Engineering ENGR 211 Instructor: K. Martin HW 1 Your final answers must be expressed with appropriate units and concern for significant figures. Problem 1: Values: W = 210 lb If a man weighs “W” on earth, specify… a) his mass in slugs b) his mass in kilograms c) his weight in Newtons If the man is on the moon, where the acceleration due to gravity is 5.30 ft/s2, determine… d) his weight in pounds e) his mass in kilograms Comments: As far as formatting is concerned, the GIVEN/FIND/SKETCH/SOLUTION format is still required for this problem, but you can make it a little more efficient since it’s not a typical workout problem. For example, a stick‐figure of a person with given weight is sufficient for the sketch. No need to repeat the sketch for each sub‐part. Kenny’s Takeaways: Recognize the difference between mass and weight. Weight is a force, and that’s what we deal with most in this class. Never report answers in units of “kg” if the question is related to forces. If the problem is in SI units, you may be given mass (kg) or force (N). If you’re given mass, you need to convert to force using “g.” In US Customary units, however, you will be given units of force (lb). No need to do any conversions – you are already in units of force. And for what its’ worth, the US Customary units of “slugs” are seldom used, so don’t worry too much about them. Use g = 9.81 m/s2 in this class. I don’t care what you did (or do) in physics. When you truncate “g” to three significant figures, it’s 9.81. Notice anything about parts (b) and (e)? Try to start thinking about stuff like this before you plug and chug with your calculator. Train your mind to look for big‐picture stuff. OREGON STATE UNIVERSITY College of Engineering ENGR 211 Instructor: K. Martin Problem 2: Values: FA = 20 N FB = 30 N α = 25° Use the graphical method to determine… a) the magnitude of the resultant force on the bolt. Use the law of cosines for this one. b) the angle made by the resultant, as measured relative to the positive x‐axis. Use the law of sines for this one. As a reminder, the figures below illustrate the graphical method. Kenny’s Takeaways: The graphical method is particularly useful when your reference axes are not perpendicular to one another This method is also great for rapidly approximating how large the resultant is and which way it is oriented. For example, if you draw a scale picture of the two forces (roughly), orient them head‐to‐
tail, and then connect them, you can see the resultant pretty quickly. The drawback here is you have to use the law of sines and cosines, which not all students are familiar with (even though you should be). Also, the graphical method gets a little cumbersome when you have more than two vectors involved. OREGON STATE UNIVERSITY College of Engineering ENGR 211 Instructor: K. Martin Problem 3: Now we will repeat the previous problem using the Cartesian method. Values: FA = 20 N FB = 30 N α = 25° Use the Cartesian method to determine… a) the magnitude of the resultant force on the bolt. b) the angle made by the resultant, as measured relative to the positive x‐axis. This is the method you are probably more familiar with. We are going to break each force into its respective x and y components, and then add the components together to get the resultant. We can therefore check our answers to the previous problem. This is what we do in the real world by the way. When we don’t have answers in the back of the book, we use different methods of achieving the same result, and if we don’t get the same result, we know something is wrong. Here’s an illustration of what we’re doing: Kenny’s Takeaways: This method is pretty clean and easy to apply, however there are some common pitfalls: o Watch your sines and cosines. If you are given angles measured from horizontal, cosine will yield x‐components as you expect. But if your angles are measured from vertical, your sines and cosines will be swapped. Watch for this – it’s a really common mistake…and totally frustrating when it happens to you. We will use this method extensively throughout this course so get familiar with it. Check your answers with the results from the previous problem. Do they jibe? They should. OREGON STATE UNIVERSITY College of Engineering ENGR 211 Instructor: K. Martin Problem 4: Values: a = 300 mm b = 75 mm c = 100 mm Use the graphical method to break F into components parallel to bars AB and BC. If the component parallel to bar AB is 7 kN, what is the magnitude of F? Comments: This is a perfect example of why the graphical method is useful – we are asked to resolve a force into components that are not perpendicular to one another. It’s difficult to apply regular old sine and cosine (i.e. trig) for this because we’re not dealing with right triangles. Instead we will use the law of sines and/or the law of cosines. Start by drawing a vertical line to represent the vector F. Now draw a line parallel to AB passing through the top of F, and another line, also parallel to AB, through the bottom. Repeat for BC. You should have a parallelogram divided into two triangles by the resultant vector F. Pick one of the triangles to work with, use geometry to figure out the angles, and then apply the law of sines to compute the magnitude of F. Remember, these aren’t right triangles so you have to use the law of sines. See below… Kenny’s Takeaways: This is a clever way to compute the force “felt” in each bar. In other words, the component of force parallel to bar AB is in fact the force in bar AB. Similarly, the force parallel to bar BC is the force in bar BC. Cool right? We will see another way of computing forces in bar assemblies like this when we study trusses later in the term. OREGON STATE UNIVERSITY College of Engineering ENGR 211 Instructor: K. Martin Problem 5: Values: FA = 110 lb FB = 120 lb θ = 15° Use the Cartesian method to compute the magnitude of the total force exerted on the pulley by the belts. Comments: Same method as problem 3. Pretty straightforward in that regard. Watch your sines and cosines! OREGON STATE UNIVERSITY College of Engineering ENGR 211 Instructor: K. Martin Problem 6: Values: A = 13 lb θ1 = 20° θ2 = 60° Use the figure shown on the right to complete the following steps: a) Express the force as a Cartesian vector. Use basic trig functions to determine the force components. Some textbooks provide equations for this, but it’s a waste of time to memorize these, or worse, to apply them without understanding what you’re doing. Just break the vector into two right triangles and apply basic trig. Nothing to memorize. Nothing fancy. The leg of one triangle becomes the hypotenuse of the other. b) Specify the coordinate direction angles of the vector. These are referred to in Hibbeler’s textbook as α, β, and γ. They are also sometimes referred to as θx, θy, and θz. See the figure below for clarification. Use vector equations for this, not trig. That is, once you have the vector components, this is a straightforward calc – no crazy trig or geometry involved. Kenny’s Takeaways: In this class we often deal with forces expressed as vectors. That’s why it’s important to be able to switch between magnitude, as given here, and the Cartesian vector notation. Sometimes you will be given α, β, and γ (as measured from the x, y, and z axes, respectively), while other times you will be given arbitrary angles. The former is obviously easier than the latter, but with that said, you don’t have to let the geometry freak you out. Look for the two triangles described above and you’re good to go. OREGON STATE UNIVERSITY College of Engineering ENGR 211 Instructor: K. Martin Problem 7: Values: h=4m a = 1.5 m b=2m FB = 400 N FC = 600 N Determine the magnitude of the resultant force on the pole. You should have one single boxed value, not a vector. As always, include units and show concern for significant figures. Comments: Here’s the general process. Position vector → Unit vector → Force vector More specifically… 1. Express the position vectors AB and AC. Pay attention to sign. When doing this, I like to think of it this way. How many steps, and in which direction, would you have to walk if you were walking from A to B and A to C? Those are your position vectors. 2. Compute the magnitude of these two position vectors now that you know their components 3. Convert each position vector to a unit vector 4. Multiply each unit vector by the force magnitude to get force vectors 5. Add the components of each force vector to get the components of the resultant. Pay attention to sign. 6. Use the components of the resultant to express the magnitude of the total force on the pole. Done. OREGON STATE UNIVERSITY College of Engineering ENGR 211 Instructor: K. Martin Problem 8: Values: d1 = 6.5 ft d2 = 2.75 ft Force in cable AB = 475 lb Use the dot product to determine the following: a) The magnitude of the force projected along the axis of the pipe (axis AC) b) The angle θ between the cable and the axis of the pipe (axis AC) Comments: For the first part of this problem you will use the dot product to compute the projection. And for this you need the force in the cable expressed as a vector. Use the conversion from position vector to unit vector to force vector as described previously. Now dot the force vector with the unit vector for the pipe. Pay attention to that. You need the unit vector for axis AC, from A to C. For what it’s worth, this represents the force “felt” along the length of the pipe as a result of the tension in the cable. Cool right? For the second part of the problem, I want you to use the dot product again. Sure, there are other ways of doing this, but I want you to be familiar with the dot product and its practical uses. Now that you have answers for both parts of this problem, can you check your values somehow? Seems to me like you could use a pretty simple equation to see if your answers jibe. That is, if you know the total force and the included angle, you can figure out the projection rather quickly... ...

View
Full Document