May The Force Be With You Worksheet

jlklk “May The Force Be With You” Alison McHugh 2/8/2024 Introduction: Forces and their behaviors form the foundation of this lab. The purpose of this lab is to enhance our comprehension of forces with a focus on identifying forces acting on objects. This lab also aims to help us understand free-body diagrams and the concept of friction. Starting with one-dimensional forces and progressing to two- dimensional forces, this lab aims to improve essential skills in physics. In the experiment, a systematic approach was used to explore forces in one and two dimensions. Methods: The general methods are described in detail in the lab manual. To reduce errors in this experiment and for the calculation of uncertainties, 4 trials were done for each section of the lab.

A: Effect of Perpendicular Forces on the Normal Force What would you expect to observe if the string were not horizontal? If the string were not horizontal, it would be expected that the aluminum block would experience a net force in the horizontal and vertical directions. Instead of moving only horizontally, the block would move diagonally. Sketch of apparatus w/ Forces Free-Body Diagram F N F x F s F g Newton’s 2 nd Law (x-direction) Σ F x = F x + F s = 0 Newton’s 2 nd Law (y-direction) Σ F y = F N + F g = 0 Would the normal force on the aluminum block change if you added additional weight to the hanger on the left? If yes, by how much? Yes, the normal force on the block would change if additional weight was added to the hanger on the left. The normal force counteracts the gravitational force acting on the object. When weight is added to the hanger, this increases the downward force acting on the entire system. Would the frictional force holding the aluminum block change if you were to add additional weight directly on top of the aluminum block? If yes, by how much? If additional weight was added directly on top of the aluminum block, the frictional force holding the block in place would increase. The increase of the frictional force would depend on various factors, but as the system reaches equilibrium would increase to match the increased normal force.

Maximum Frictional Force Trial Horizontal Hanging Mass (g) Maximum Static Frictional Force (N) Vertical Hanging Mass (g) Normal Force (N) 1 150 1.47 90 3.397 2 250 2.45 55 3.361 3 150 1.47 75 3.397 4 160 1.57 76 3.318 Calculate an estimated coefficient of static friction between the aluminum block and the balance with its absolute uncertainty. 𝜇 𝑠 = (0.4327 + 0.72894 + 0.432734 + 0.47318) / 4 = 0.5169 Σ (F friction / F N ) / N δ 𝜇 𝑠 = σ 𝜇 𝑠 / √ N = 0.1235 / √ 4 = 0.06175 C: Forces in Multiple Directions Sketch of apparatus w/ Forces Free-Body Diagram F N F F x F s F g Newton’s 2 nd Law (x-direction) Σ F x = F x + F s + F 30x = 0 Newton’s 2 nd Law (y-direction) Σ F y = F N + F g + F 30y = 0

Maximum Frictional Force Trial Horizontal Hanging Mass (g) Maximum Static Frictional Force (N) Vertical Hanging Mass (g) Normal Force (N) 1 101 0.9898 150 3.476 2 99 0.9702 152 3.464 3 95 0.931 156 3.407 4 97 0.951 154 3.398 Compared to Part B, do you need more or less weight on the horizontal hanger to cause the weight to move? Do your observations make sense? Compared to Part B, you would need less weight on the horizontal hanger to cause the weight to move in Part C. Since sin(30) is around 0.5, you would need half the weight on the horizontal hanger compared to Part B to overcome static friction. Calculate the estimated coefficient of static friction with an absolute uncertainty. µ s = F s ,max − verticle× sin ( 30 ) N 𝜇 𝑠 = (0.2845 + 0.2801 +0.2733 + 0.2799) / 4 = 0.2795 Σ (F friction / F N ) / N δ 𝜇 𝑠 = σ 𝜇 𝑠 / √ N = 0.004 / √ 4 = 0.02

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