D] A 4 kg mass in free space (no gravity) is seen moving with a constant acceleration of 2 m/s² in the +j direction because of four forces, three of which are known and shown. Using Fnet = F₁+F2 + F3 = ma, find the unknown force F4 in unit vector notation. +y |F1|=10 60° F₂ 120 +X F₁ = F₂ = F3 = -5 -10 0 F₁+F₂+f3+fu=8 *15 78-Fu i+ 8.66 i+ 0 i+ -10 Fu=23-7 F12 = -15 1+ 41.34; F₁ = 15 1+ 1.34; COE Fnet i+ -10 目
D] A 4 kg mass in free space (no gravity) is seen moving with a constant acceleration of 2 m/s² in the +j direction because of four forces, three of which are known and shown. Using Fnet = F₁+F2 + F3 = ma, find the unknown force F4 in unit vector notation. +y |F1|=10 60° F₂ 120 +X F₁ = F₂ = F3 = -5 -10 0 F₁+F₂+f3+fu=8 *15 78-Fu i+ 8.66 i+ 0 i+ -10 Fu=23-7 F12 = -15 1+ 41.34; F₁ = 15 1+ 1.34; COE Fnet i+ -10 目
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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![0] A 4 kg mass in free space (no gravity) is seen moving with a constant acceleration of 2 m/s² in the
+j direction because of four forces, three of which are known and shown.
Using Fnet = F1+F2 + F3 = ma, find the unknown force F4 in unit vector notation.
+y
|F1 = 10
F123 = -15
F4=15
60°
F2
(
+X
F₁=5
F₂ = -10
F3 =
0
F₁+√₂+f3+fu=8
*15
78-Fy
i+ 8.66
i+
0
i+
-10
Fu = -7
Fnet
_i+ -1.34;
11+ 1.34 COE
i+
-10
(0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa1d592a5-ec96-4b1d-bddf-23e27014debe%2Fb94929c9-b676-4cb1-b696-f1adefc92a20%2Fvcy0j2b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:0] A 4 kg mass in free space (no gravity) is seen moving with a constant acceleration of 2 m/s² in the
+j direction because of four forces, three of which are known and shown.
Using Fnet = F1+F2 + F3 = ma, find the unknown force F4 in unit vector notation.
+y
|F1 = 10
F123 = -15
F4=15
60°
F2
(
+X
F₁=5
F₂ = -10
F3 =
0
F₁+√₂+f3+fu=8
*15
78-Fy
i+ 8.66
i+
0
i+
-10
Fu = -7
Fnet
_i+ -1.34;
11+ 1.34 COE
i+
-10
(0
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