Critical Analysis Of Load Distribution On a scissor Jack Gabriel Chong, Paul Stegner, Chris Yorgey Department of Physics and Engineering Elizabethtown College, Elizabethtown, PA Email: [email protected] edu, [email protected]

edu, [email protected] edu INTRODUCTION In this discussion we will introduce our critical analysis of the load distribution on a scissor jack. It has been specified that this jack is to hold a load of 100 lbs on the top plate, with the jack height set to 7 inches. Our analysis will focus in on the forces experienced by the screw which adjusts the height of the scissor jack.We will solve the problem using general formulas and variables so that our results can be universally applied. With the given information, we will determine the load as it is redistributed from the plates to the legs, and from the legs into the pegs that act as supports. We hypothesize that by solving the generic system for this scissor jack by assuming the weight is evenly distributed, we can gain better knowledge of the interaction of forces on the scissor jack, including the effects of placing the load in different positions. PROCEDURES Our main goal as team A.

W. E. S. O.

M. E. (Always Working Extremely Serious On Momentous Experiments) was to solve the system based on a generic system of equations. We accomplished this task by assuming that the 100 lb force is being evenly distributed on all four pins. Knowing this, we could easily solve the system by viewing the FBD’s in two dimensions. Thus, in order to get the correct value, all we would have to do is multiply our theoretical answer by 2 to include the opposite side of the 2-D diagram.

By solving for variables before using any numbers, we simplified our calculations.RESULTS We used the fundamental equations below for all our calculations: | |? [pic] |(1) | | |[pic] |(2) | [pic] Figure 1: Free Body Diagram of the Lab Jack [pic] Figure 2: Free Body Diagrams of Forces on Individual Points From Figure 2 we were able to find the following equations. At points A and C (where [pic] is replaced with [pic] for point C): | |? [pic] |(3) | | |? pic] |(4) | At point B: | |? [pic] |(5) | | |? [pic] |(6) | Equation 6 tells us that [pic] is equivalent to[pic]. From this fact we were able derive the following equation for the force on the pin: | |[pic] |(7) | Next we sought to describe theta, as seen in Figure 1, in terms of the length of each cross-brace, d, and the height, h.We found the following equation: | |[pic] |(8) | From Equation 4 we are able to derive and equation for [pic] in terms of the force applied and the angle, theta, as described in our free body diagram. Combining this with our knowledge of theta from Equation 8 we found the equation we are able to put Equation 7 in terms of the major variables pertaining to this problem.

| |[pic] |(9) |We then got rid of the trigonometry involved and accounted for the fact that this diagram is symmetric and the other side would be experiencing equal forces to find the total force on the pin. | |[pic] |(10) | From this result we are able to calculate the force experienced by the screw given any height to which the jack is raised, any applied force and any scaled jack with different cross-brace lengths. By plugging in the values [pic]=100 lbs, d = 4[pic]” and h = 7” we found that the force on the pin was around 86. 4 lbs. DISCUSSION AND SUMMARYWe feel our final result is both reasonable and accurate. Looking at the impracticality of some of the simplifying assumptions we made in our analysis, such as a uniformly distributed load, we began looking at the implications of different load distributions. If we were to move the load to another other spot on top of the scissor jack, the total weight of the load will still be proportionally distributed amongst the legs and the screw within the stand. In essence, changing the load point of reference on the top of the scissor jack would have very little effect on the numerical values we calculated.

REFERENCES Hibbeler, Statics, Pearson 2007