The equations are in the terms of the variables given in the problem a) Draw a Free Bady Diagramb) Write an eguation for net for in the x-directionFnet, x =%3D)Write an eguatien for net force in the Y-directionFnet, Y= Problem 4: A block with a mass of m = 39 kg rests on a frictionless surface and issubject to two forces acting on it. The first force is directed in the negative x-direction witha magnitude of F1 = 11.5 N. The second has a magnitude of F2 = 21 N and acts on the bodyat an angle 0 = 21° measured from horizontal, as shown.F2

Answered: a) Draw a Free Bady Diagram b) Write an…

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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The equations are in the terms of the variables given in the problem

a) Draw a Free Bady Diagram
b) Write an eguation for net for in the x-direction
Fnet, x =
%3D
)Write an eguatien for net force in the Y-direction
Fnet, Y=

Problem 4: A block with a mass of m = 39 kg rests on a frictionless surface and is
subject to two forces acting on it. The first force is directed in the negative x-direction with
a magnitude of F1 = 11.5 N. The second has a magnitude of F2 = 21 N and acts on the body
at an angle 0 = 21° measured from horizontal, as shown.
F2

Expert Solution

Step 1

A force is an interaction of a push or pull action that occurs to any object. One force always acting on all bodies on Earth is the force of gravity. This force is directed towards the Earth's center.

A free-body diagram is a figure that shows all the force that is currently working on an object and in which direction.

Step 2

The free-body diagram is given below.

Considering the positive x and y-direction as positive, and the negative x and y-direction as negative.

Using the FBD we have.

The net force in the x-direction is

$F_{net, x} = F_{2} cosθ - F_{1}$

Putting the values we have

$\begin{array}{rcl} F_{net, x} & = & F_{2} cosθ - F_{1} \\ = & 21 N \cos 21^{0} - 11.5 N \\ = & 8.11 N \end{array}$

The net force in the y-direction is

$F_{net, y} = F_{2} sinθ + mg$

Putting the values we have

$\begin{array}{rcl} F_{net, y} & = & F_{2} sinθ + mg \\ = & 21 N \sin 21^{0} + 39 kg \times 9.8 m / s^{2} \\ = & 389.73 N \end{array}$

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