Test Information Description PHY 1100 Lab 4: Atwood's Machine Instructions Note: You are expected to format all variable

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Test Information Description Phy 1100 Lab 4 Atwood S Machine Instructions Note You Are Expected To Format All Variable 1 (540.5 KiB) Viewed 51 times

Test Information Description Phy 1100 Lab 4 Atwood S Machine Instructions Note You Are Expected To Format All Variable 2 (338.17 KiB) Viewed 51 times

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Test Information Description PHY 1100 Lab 4: Atwood's Machine Instructions Note: You are expected to format all variables/equations using the math editor which you access by clicking on the square root symbol on the toolbar. Equipment Needed 1 Photogate/Pulley System (ME-6838) 1 Mass and Hanger Set (ME-8967) 1 Braided Physics String (SE-8050) 1 Universal Table Clamp (ME-9376B) Background The acceleration of an object depends on the net applied force and the object's mass. In an Atwood's Machine, Mass 1 Pulley Mass 2

assuming that 1. the pulley is massless and frictionless and i. the string is massless and does not stretch, the difference in weight between the two hanging masses determines the net force acting on the system of both masses. This net force accelerates both of the hanging masses; the lighter mass M₁ is accelerated upward and the heavier mass M₂ is accelerated downward. The magnitude Fnet of this net force can thus be written Fnet S (M₂ - M₁) g where is the magnitude of the acceleration of the system. It follows that (M₁ + M₂) a, (M₂-M₁) g M₁ + M₂ 1 We can derive this result by considering the following freebody diagrams. (2)

We can derive this result by considering the following freebody diagrams. T₁ M₁ M₁ g 1 T₂ M₂ M₂9 Here, T₁ is the magnitude of the tension in the string on the side attached to M₁ and ₂ is the magnitude of the tension in the string on the side attached to M₂. Taking the convention that up is positive and down is negative, we obtain F₁ = T₁ - M₁ g = M₁ ª₁ = M₁ a. F₂ = T₂ - M₂ g = M₂ Q₂ = M₂ (–a), where F₁ and F₂ are the vertical components of the net forces on M₁ and M₂, respectively. Subtracting (4) from (3) (3) (4)

Here, T₁ is the magnitude of the tension in the string on the side attached to M₁ and T₂ is the magnitude of the tension in the string on the side attached to M₂. 1 Taking the convention that up is positive and down is negative, we obtain where F1 F₁ = T₁-M₁ g = M₁ a₁ = M₁ a, F₂ = T₂ M₂ g = M₂ a₂ = M₂ (-a), and F₂ are the vertical components of the net forces on M₁ and M₂, respectively. Subtracting (4) from (3) (5) T₁ T₂ + (M₂ - M₁) g = (M₁ Using (6) in (5) yields (2). = The assumptions above about the pulley and string ensure that 1. Ain T₁ = T₂. + M₂) a. Save and Submit