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Prathik el x VA 201 Lab5 -Torques & S x G The line that is the best fit x C fi w

ID: 1343313 • Letter: P

Question

Prathik el x VA 201 Lab5 -Torques & S x G The line that is the best fit x C fi www.webassign.ne 11803076 t/web/Student/Assignment-Responses/submit?dep EE: Apps Bookmarks ported From IE owl V2 Cengage L. WA WebAssign LOGIN NA eCampus Learning Texas A&M; Universi... P Active Courses-Pear... Assignments Other bookmarks A uniform metal bar is 5.00 m long and has mass 0.300 kg. The bar is pivoted on a narrow support that is 2.00 m from the left-hand end of the bar. What distance x from the left hand end of the bar should an object with mass 0.900 kg be suspended so the bar is balanced in a horizontal position? 1.834 m Additional Materials L Torques and Static Equilibrium 0/10 points I Previous Answers TAMUColPhysMechL1 5.POST.003 My Notes Ask Your Teacher The experiment you did in lab is repeated, using a uniform metal bar that is 80 cm long instead of the stick. Since the bar i uniform, its center of gravity is at its center The new experiment uses different hooks for hanging the masses from the bar, with mhook 5.0 g. As in the experiment you did in lab, x1 5.00 cm m1 300.0 g, and xp. 25.0 cm In the new experiment, you make the same measurements as in your lab and plot x versus m I The line that is the best fit to your data has slope m hook 3600 cm g. What is the mass of the bar? 200 g Additional Materials L Torques and Static Equilibrium Submit Answer Submit Assignment Home My Assignments Extension Request

Explanation / Answer

2. balancing moment(r x F) about hinge point,

(2.5 - 2) x 0.30 x 9.80 - (d x 0.900 x 9.8) = 0

d = 0.15 / 0.9 = 0.167 m

distance from left end = 2 - 0.167 = 1.83 m


3 . data from experiment needed.

figure etc.


3. F = mg = kx

0.030 kg x 9.80 m/s^2 = 8.80 N/m x X

X = 0.0334 m Or 3.34 cm


5. F = kx

16 = k (0.0590)

k = 271.19 N/m

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