2012 ppt hanging mass
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1) The document instructs the reader to draw a force diagram of an object suspended by two spring scales, using four steps: center of mass, long term forces, contact forces, and labeled axes. 2) The force diagram shows two tension forces (T1 and T2) acting at angles, with T1 measured at 1N. 3) Calculations show that T2 must also be 1N in the x-direction and 4.69N in the y-direction, in order for the object to be in translational equilibrium.
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Here are the steps to solve this problem:
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Use: vf = vi + at
Substitute: vf = 0 + (0.21 m/s2)t
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Solve for t: t = √(280/0.105) = 60 s
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The document describes an experiment to determine the natural frequency of a spring-mass system. A spring is attached vertically to an adjustable screw and platform. Weights are added to the platform to stretch the spring from its free length. The time for a set number of oscillations is measured for different weights. This is used to calculate the experimental periodic time and natural frequency, which are compared to the theoretical values calculated using the spring constant and mass. The results show that the experimental and theoretical natural frequencies match closely.
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Https _www.math.umass.edu_~tanguay_331_331spr10hw17sol
Https _www.math.umass.edu_~tanguay_331_331spr10hw17sol
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2012 ppt hanging mass
- 3. Draw a force diagram of the object suspended from two spring scales that is set up on the front table. Remember to use the four steps (C of M, long term forces, contact forces, labeled axes) in creating a force diagram!
- 4. y T2 78° T1 (1N) x Fg
- 5. Because we are in translational equilibrium, we know that: T1 = T2x T2 COS 78° = T2x (T2) COS 78° = 1 N (T2) (0.208) = 1 N T2 = 1 N/0.208
- 6. T2 SIN θ = T2y 4.8 N SIN 78° = T2y (4.8 N) (0.978) = T2y 4.69 N = T2y
- 7. y T2y 4.69 N x T1 1 N T2x 1 N Fg
- 8. Fg = may and T2y = may 4.69 N = may 4.69 N = (m)(9.8 m/s2) 4.69 N/9.8 m/s2 = (m) 4.69 N/9.8 m/s2 = .478 kg Now, lets put the object on the balance!