圖: http://i187.photobucket.com/albums/x170/awial/phyk.jpg
On a horizontal ground, Peter of mass 55kg sits on an office chair of 5kg and both of them are initially at rest. At the instant shown (上圖), A force of 200N is applied at an angle of 50度 to the horizontal on the chair.
Find the magnitude of force exerted on Peter by the chair.
ANS: 565N
f.4 phy-力學
2007-06-18 2:03 am
回答 (2)
2007-06-18 2:26 am
✔ 最佳答案
圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/phyk2.jpg?t=1182075747
The resultant force exerted on Peter can be resolved into 2 forces, one vertical and one horizontal.
For the vertical component, reaction exerted on Peter by the chair
= weight of Peter
= mg
= 55 X 10
= 550 N
For the horizontal component, the force which is responsible for the acceleration for Peter, F
= 200cos50°
= 129 N
So, the magnitude of force exerted on Peter by the chair
= √[(129)² + (550)²]
= 565 N
參考: Myself~~~
2007-06-18 2:27 am
Before the 200N force is applied, Peter is acted on by the upward normal reaction (which equals to his weight) given by the chair, i.e. 550 N.
The horizontal component acting on the chair by the 200N force
= 200xcos(50) N = 128.56 N
The force will be given by the chair and acts on Peter in the horizontal direction.
Hence, the resultant force acting on Peter
= square-root[128.56^2 + 550^2] N
= 565 N
The horizontal component acting on the chair by the 200N force
= 200xcos(50) N = 128.56 N
The force will be given by the chair and acts on Peter in the horizontal direction.
Hence, the resultant force acting on Peter
= square-root[128.56^2 + 550^2] N
= 565 N
收錄日期: 2021-04-29 17:24:21
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070617000051KK03181