A 500 gram meter stick has a center of mass at the 50 cm mark. Using two fingers, one at the 20 cm mark and the other at the 70 cm mark, you hold the meter stick in a horizontal, stationary position. Find the forces your fingers place on the meter stick. Four point masses, m1 = 1 kg, m2 = 2 kg, m3 = 3 kg and m4 = 4 kg are on the x axis as shown in the figure below. Find the moment of inertia of the above system about an axis of rotation that is perpendicular to the page and passes through x = 0 m. Find the moment of inertia of the above system about an axis of rotation that is perpendicular to the page and passes through x = 4 m. 7. A 500 gram meter stick has a center of mass at the 50 cm mark. Using two fingers, one at the 20 cm mark and the other at the 70 cm mark, youhold the meter stick in a horizontal, stationary position. Find the forces your fingers place on the meter stick.Four point masses, mj = 1 kg, m2 = 2 kg, m3 = 3 kg and m4 = 4 kg are on the x axis as shown in the figure below.mim2m3m46.х (m)8. (Extra Credit) Find the moment of inertia of the above system about an axis of rotation that is perpendicular to the page and passes through x= 0 m.9. (Extra Credit) Find the moment of inertia of the above system about an axis of rotation that is perpendicular to the page and passes through x= 4 m.00

7) Given the mass of the stick is m = 500 g = 0.5 kg center of mass of the stick is at x = 50 cm and the stick is supported at the points x = 20 cm and x = 70 cm so by applying the moment equilibrium the net moment about the point x = 0 will be zero as the stick is not rotating so Mnet = F20(0.2 m)-mg(0.5 m)+F70(0.7 m) 0 =0.2F20-(0.5 kg)(9.8 m/s2)(0.5 m)+0.7F70 2.45 =0.2F20+0.7F70 ...1 Now by the Linear equilibrium as the stick is linearly stable so the net upward force equals the net downward force F20+F70 = mg =0.5*9.8F20+F70 = 4.9 ...2 Now using the equations 1 and 2 we get: F20 + F70=4.9 0.2F20 + 0.7F70=2.45 0.2(4.9-F70) + 0.7F70=2.45 F70 = 2.94 NNow by the equation number 2 F20 =4.9 -2.94F20 =1.96 N So the force on the fingers will be different and equals F20 = 1.96 N ; F70 = 2.94 N Thank you!

Answered: 7. A 500 gram meter stick has a center…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A 500 gram meter stick has a center of mass at the 50 cm mark. Using two fingers, one at the 20 cm mark and the other at the 70 cm mark, you hold the meter stick in a horizontal, stationary position. Find the forces your fingers place on the meter stick.

Four point masses, m₁ = 1 kg, m₂ = 2 kg, m₃ = 3 kg and m₄ = 4 kg are on the x axis as shown in the figure below.

Find the moment of inertia of the above system about an axis of rotation that is perpendicular to the page and passes through x = 0 m.

Find the moment of inertia of the above system about an axis of rotation that is perpendicular to the page and passes through x = 4 m.

7. A 500 gram meter stick has a center of mass at the 50 cm mark. Using two fingers, one at the 20 cm mark and the other at the 70 cm mark, you
hold the meter stick in a horizontal, stationary position. Find the forces your fingers place on the meter stick.
Four point masses, mj = 1 kg, m2 = 2 kg, m3 = 3 kg and m4 = 4 kg are on the x axis as shown in the figure below.
mi
m2
m3
m4
6.
х (m)
8. (Extra Credit) Find the moment of inertia of the above system about an axis of rotation that is perpendicular to the page and passes through x
= 0 m.
9. (Extra Credit) Find the moment of inertia of the above system about an axis of rotation that is perpendicular to the page and passes through x
= 4 m.
00

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