Mr. One is pushing a cart with a force of F1 = 15.00 N due north. Mr. Two is pushing the same cart with a force of F2 = 25.00 N due northeast. Mr. Three is pushing the ca a force F3 so that the cart is in equilibrium. Find force F3 so that F1 + F2+F3=0.

Mr. One is pushing a cart with a force of F1 = 15.00 N due north. Mr. Two is pushing the same cart with a force of F2 = 25.00 N due northeast. Mr. Three is pushing the ca a force F3 so that the cart is in equilibrium. Find force F3 so that F1 + F2+F3=0.

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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Concept explainers

Newton’s Third Law of Motion

The horse pulls the cart and the cart pulls the horse in response as per Newton third law. So the big question is, “Don’t the two forces cancel each other? How does the cart accelerate?”

Question

Supply all missing information with the correct numerical values. Do not include the units. Round off all answers to two decimal places. Do not forget the
negative sign (-) if needed.
Mr. One is pushing a cart with a force of F1 = 15.00 N due north. Mr. Two is pushing the same cart with a force of F2 = 25.00 N due northeast. Mr. Three is pushing the cart
a force F3 so that the cart is in equilibrium. Find force F3 so that F+F,+F3=0.
First, we calculate the vector sum
+ F2 by component method.
The x-component of F1 is
and the y-component is
The x-component of F2 is
and the y-component is
Adding the components, we can express the resultant vector in unit vector component form:
Fi + F2 =
Î +
Since we want to find Fj + F,+ F3 = 0, it follows that
F3 = - (F+F2). Thus.
F3=
The magnitude of F3 is F3 =
N.

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