Use the principle of equilibrium (where the sum of torques is zero) to calculate the third, unknown force acting at x3 (what is m3)? 620010509030653.103.31 Trialm₁ (g)X₁ (cm)m₂ (g)X2 (cm)X3 (cm)fulcrum location (cm)Minimummeasured F3 (N)Maximummeasured F3 (N)

The principle of equilibrium suggests that the net torque about fulcrum must be zero. The FBD of the…

Answered: Use the principle of equilibrium (where…

Use the principle of equilibrium (where the sum of torques is zero) to calculate the third, unknown force acting at x3 (what is m3)?

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter1: Physics And Measurement

Section: Chapter Questions

Problem 1.2OQ: A house is advertised as having 1 420 square feet under its roof. What is area in square meters? (a)...

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Concept explainers

Rotational Equilibrium And Rotational Dynamics

In physics, the state of balance between the forces and the dynamics of motion is called the equilibrium state. The balance between various forces acting on a system in a rotational motion is called rotational equilibrium or rotational dynamics.

Equilibrium of Forces

The tension created on one body during push or pull is known as force.

Question

Use the principle of equilibrium (where the sum of torques is zero) to calculate the third, unknown force acting at x₃(what is m₃)?

Trial
m₁ (g)
X₁ (cm)
m₂ (g)
X2 (cm)
X3 (cm)
fulcrum location (cm)
Minimum
measured F3 (N)
Maximum
measured F3 (N)

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w = FjAc = F .Aë , W = m(v;)* - m(u,)* , v 1 zm(us)* - m(v,)* , W = Problem 1: (a) Ã = 3â – 2ŷ, B = 4â& – 5ŷ. Ã ·B =? Answer: 22 (b) Ã = -2â + 4ŷ, B = 3âu – lŷ. Ā·B =? Answer: –10 (c) Ã = 5â – 3ŷ, B = -3â – 5ŷ. Ã·B =? Answer: 0 1 | F,dx , W = F. dř , F, dU W = -AU dx d = 4m F = 20 N 0 = 30° m = 4 kg Hk = 0.5) m = 2 kg F = 40 N Hk = 0.2 Ө — 30° Problem 2: In the figure above on the left we have a force F = 20 N pulling a box to the right at an angle 0 = 30° above the horizontal. The box has a mass m = 2 kg, the coefficient of kinetic firction between the box and the ground is uk = 0.2, and the block moves a distance d = 4 m to the right. (a) What is the work done by the force F? Answer: 69.3 J (b) What is the normal force from the ground? Use g = 10 m/sec² Answer: 10 N (c) What is the work done by the kinetic friction force? Answer: -8 J (d) What is Wnet, the net work done on the box? Answer: 61.3 J (e) The box has an initial speed of 4 m/sec. What is the speed of the box after it has…