The solutions to the problems below can be found at the end of this page. Try all the problems before looking at the solutions. It's much easier to understand a solution put before you than to come up with the solution yourself. To develop the skills necessary to solve the problems yourself, you must spend the time doing it.
Problems
You want to build a storage shelf in the garage. It is to be used for paints, oil for the car, bleach, laundry detergent, etc. You decide that you want it to be a foot wide and ten feet long. This way you can store two rows of larger items such as bottles of bleach or cans of paint. First we need an estimate for the total weight it is expected to hold. A gallon of water weighs eight pounds. A bleach bottle or a can of paint is about 9 inches wide. So, we might have a total weight of (10 ft x 12 in / 9in) x 2rows x 8 lb = 210 lb approximately. The shelf will be rest on a runner screwed securely into the wall. The outside edge of the shelf is to be supported with a nylon cord that is safety-rated to a tension of 15 lb. The twine will hang vertically downward from screweyes in the ceiling. How many strings should you use?
You've decided to build a new deck. It is to be 10 ft wide and run the 30 ft length of the house. You must estimate the weight of the deck, as well as, the maximum load you expect it to support. That means you need an estimate of the weight of the lumber and how many people you expect at your deck party. A reasonable value might be around 6,000 lb. For now, lets denote the total load as W, assumed to be spread uniformly over the deck. The deck is to be supported where it is attached to the house and at the outside edge by support beams that make a 45o angle as shown. What will be compressive tension in the beams?
Installing new cabinents can really spruce up the look and functionality of your kitchen. But what kind of attachments are necessary? A standard method of installation is to attach a running support strip along the wall for the bottom edge of the cabinents to rest upon. The bottom strip supports the weight of the cabinets. The cabinents are held against the wall with screws at the top of the cabinents that screw directly into the wall. (For typical wood frame and drywall construction, it is very important to make sure the screws go into studs and not just the drywall. For concrete walls, you will need to install something like anchor bolts.) Assume the cabinets are 1 ft deep by 3 ft high and will weigh 200 lb (uniformly distributed) when full of dishes. What total horizontal force is required of the top screws? What force is supplied by the bottom strip? For the bottom strip, break the force into two forces, one horizontal and the other vertical as shown.
Solutions
Don't look at the answers until you've tried the problems on your own!!

S M = +(3 kg x 9.8 m/s2) x - (1 kg x 9.8 m/s2) (1.0 - x) = 0
Solve for x to find x = .25 m or 25 cm. Notice that the force exerted by the string does not appear in the equation since the distance from the pivot is zero!
Look at the diagram. Using the hints given, we draw all the forces acting. We take the pivot point at the bottom of the rafter. Since we have an equilateral triangle, the moment arms are easy to figure out. Directly below the point where the two rafters meet must be exactly half the distance, or 4 ft along the joist. So the line of action of the weight (acting at the center of the rafter) must be 2 ft. Using the Pythagorean theorem, the line of action of the force at the top must be
(82 - 42) = 6.9 ft. Noting the directions, the moments equation gives us S M = +100 lb x 2 ft - Ftop x 6.9 ft = 0
Note that by choosing the bottom as our pivot, the bottom components don't appear in the equation. Solving for the force at the top, we have Ftop = 29 lb. Uisng the sum of forces equal to zero, it's easy to find that the force components at the bottom are Fhoriz = 29 lb and Fvert = 100 lb.
Solving for T = 210x6/12 = 105 lb. The number of cords needed is 105/15 = 7. That a cord every 20 inches. And that's the minimum. A good engineering rule of thumb is a safety factor of two. So double the number, or better yet get stronger cord.
You are now in a position to find the force supplied by the wall. The sum of forces must be zero. Can you determine that the force exerted by the wall support must be 105 lb directed upwards? This information might be useful in decide how big of screws to use.
How do you find the perpendicular distance? It depends upon your level of mathematical expertise. (You can always lay sticks and strings out on a table and measure it directly.) The length of the cord from the Pythogorean Theorem is the square root of 12 + 32 = 3.16. If you know your trigonometry, you can use sines and cosines. But here an interesting geometry trick. The area of a triangle is one-half of the base times the height. Taking the 1 ft shelf as the base and the 3 ft as the height, we get 1.5 ft2. Now take the hypotenuse (the cord) as the base and d as the height. So 1/2 (3.16)d = 1.5 ft2, and d = .95 ft or about 11 inches. Plugging into the moments equation, the new tension will be 110 lb. So we'll need 7.4 cords, instead of 7. Not a big difference if we plan to double for the safety factor anyway. However, as the wall attachment of the cord is moved closer to the shelf, the tension needed grows quickly. If we attach the strings at only 1 ft above the shelf, the tension needed is 40% greater than before. At 6 inches, the tension is over 220 lb (> 200% increase!)
This problem is quite similar to that above. We take our pivot point at the house. Assume the load, w, acts at the center of the deck. The perpendicular distance, d, to the support beams can be found using a scaled drawing or using the same trick as in the previous problem. Using the area of the triangle formed by the beam and deck, see if you can show that the beams must be 14.1 ft long and that d = 7.1 ft. The moments equation is
giving a value of T = .70 W. That is, the beams must support about 70% of the total weight load. Using load factor tables for standard lumber, you can decide how many support beams to use and how big they should be. And don't forget the safety factor!
F is the force required of the upper screws and w is the weight. The force at the bottom strip has been broken into horizontal (Nhor)and vertical (Nvert) components. We take the pivot at the bottom strip. We get
Which gives a total force required from the screws of F = 33 lb. The force(s) supplied by the bottom strip can be found by requiring the total force to be zero. But don't forget force has direction. The total force in the vertical must be zero and the total force in the horizontal must be also be zero. We write S F = 0 separately for both directions.
S Fhor = - Nhor - 33 lb = 0
The first equation gives Nvert = 200 lb. Clearly, the support strip should be securely fastened to the wall. The force on the screws will be shear forces and any reasonably sized set of screws should be able to hand this.
the second equation gives Nhor = -33 lb. What does the minus sign mean? It means we chose the wrong direction for Nhor in the picture. It actually points away from the wall. That's right, the wall is pushing out on the cabinents at the bottom! Are you suprised? Installing screws at the bottom (a good idea, by the way) serves only to insure that the cabinents are not accidently pulled off the support strip when someone slides a heavy plate out of the cabinent.
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