Static Equilibrium Problems

Here's a few problems that you might encounter when installing a shelf, making structural repairs, or building a house from scratch. Regardless of the complexity or size of the task, it's good to know the basics of static equilibrium.

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

  1. A 3 kg block rests on a horizontal table. What are the magnitude and direction of all the forces acting on the block?

  2. Place a 2 kg block on top of the 3 kg block in the previous example. What are the magnitude and direction of all the forces acting on the 3 kg block?

  3. You are making a mobile. You have two pieces with masses 1 kg and 3 kg respectively. You hang them from either end of a 1.0 m long support rod. Where must you place the suspending string (pivot) to assure static equilibrium? (Hint: Call the distance from the pivot to one of the masses "x". Then require the sum of moments to be zero.)

  4. Two rafters and a joist, all 8 ft in length, form an equilateral triangle. Once the roof is completed, each rafter must support a vertical weight of 100 lb. What are the forces acting at each end of either rafter? (Hint: If the roof is constructed symmetrically, Newton's third law says the force at the top of either rafter must be horizontal. The 100 lb will act at the center of the rafter. Break the bottom force into components. Now require the sum of the moments to be zero and then the sum of the forces to be zero)

  5. 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?

  6. Let's push this problem a bit more. What if you can't tie straight up into the ceiling. Instead, you want to tie the cords into wall, a distance of 3 ft from the shelf. Now the perpendicular distance from the pivot to the line of action of the cord tension isn't 12 inches! Can you find the perpendicular distance? Try it! If you don't have the mathematical skills to calculate it directly, get an estimate by making a scaled measurement of the diagram.

  7. A Deck

    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?

  8. Kitchen Cabinets

    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!!

  1. The weight of the block is w = mg = 3 kg x 9.8 m/s2 = 29 N and acts downward. The only other object effecting the block is the table, N. Since the total force must be zero, N - 29 N = 0 and the table must be pushing upwards with a force exactly equal to the weight, 29 N.

  2. First, we must analyze the forces on the 2 kg block. That procedure is exactly the same as above. The 3 kg block must pushing upward on the 2 kg block with a force equal to the 2 kg block's weight = 2 kg x 9.8 m/s2 = 20 N. But Newton's 3rd law states that the 2 kg block pushes with an equal but opposite force of 20 N downward. So, there are now three forces acting on the 3 kg block. Applying the sum of forces equal zero, we have N - 20 N - 29 N = 0, and so N = 49 N. as you might have guessed, this is just the total weight of the two blocks.

  3. The forces acting on the rod are the weights of each piece and the force of the supporting string. We want the string to be attached at the pivot. The distance from the pivot to the 3 kg piece is x. So the distance to the 1 kg piece must be 1.0 - x. The 3 kg piece tends to rotate the mobile cw, so its moment is positive, while the 1 kg tends to rotate ccw, or negative. The moments equation looks like

    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!

  4. 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.

  5. Look at the diagram. We estimate that the average weight will act at the center of the shelf, 6 inches from the wall. Let w = 210 lb and let T represent the total tension force needed from the cord. If we take the pivot at the wall (that way we don't need to find the forces there) we demand that the moments sum to zero. So

    S M = +(210lb)(6in) - (T)(12in) = 0

    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.

  6. 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!)

  7. A Deck

    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

    S M = +(W)(5 ft) - (T)(7.1 ft) = 0

    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!

  8. Kitchen Cabinets

    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

    S M = +(200 lb)(6 in) - (F)(36 in) = 0

    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 Fvert = Nvert - 200 lb = 0

    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.

RETURN