Use your free body diagram above and apply Newton’s second law to express the normal force in terms of the mass of the crate, the gravitational constant, and the angle of inclination. m = 100 kgg = 9.8 m/s²θ = 30° Free Body DiaO Show O HideFriction-FNO Ice (no friction)O Woodh= 5.0mVectors-V Force VectorsO Sum of Forces-10ÖmetersFree Body Diagram2000.0Anale = 30.0 *Parallel Forces (N)FApplied1000.0FFriction= 254.611 NFGravity O =-490 NFWallFN0.0= ONFSum O= -235.389 N-1000.0-2000.06101214time (sec)O RecordApplied Forceరా

Answered: Image /qna-images/answer/b8d20bd7-892e-452d-b960-dbc1b2b5a5ef.jpg

Use your free body diagram above and apply Newton’s second law to express the normal force in terms of the mass of the crate, the gravitational constant, and the angle of inclination. m = 100 kg g = 9.8 m/s² θ = 30°

Use your free body diagram above and apply Newton’s second law to express the normal force in terms of the mass of the crate, the gravitational constant, and the angle of inclination. m = 100 kg g = 9.8 m/s² θ = 30°

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter7: Gravitation

Section: Chapter Questions

Problem 57A

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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

Free Body Dia
O Show O Hide
Friction-
FN
O Ice (no friction)
O Wood
h= 5.0m
Vectors-
V Force Vectors
O Sum of Forces
-10
Ömeters
Free Body Diagram
2000.0
Anale = 30.0 *
Parallel Forces (N)
FApplied
1000.0
FFriction
= 254.611 N
FGravity O =-490 N
FWall
FN
0.0
= ON
FSum O
= -235.389 N
-1000.0
-2000.0
6
10
12
14
time (sec)
O Record
Applied Force
రా

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