MAE 101 homework 3 sol

# MAE 101 homework 3 sol - MAE 101: Statics and Strength of...

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MAE 101: Statics and Strength of Materials Fall 2011 Problem 6. (Problem 2.17 from the Gross et al.) A smooth Sphere (weight G , radius R ) rests on three points A , B and C . These three points form an equivaleteral triangle in a horizontal plane. The height of the triangle is 3 a = 3 4 3 R (see Figure). The action line of the applied force F passes through the center of the sphere. Determine the contact forces at A,B and C . Find the magnitude of the Force F required to lift the sphere off at C . Solution: Let us assume the Coordinate system is positioned at the center of the sphere. Let us call the center of the Sphere as O . We will now write the coodinates of the points AB and C . It is given that points AB C form an equilateral triangle. Let the length of each side by s . We are given CP = 3 a . Hence we have AC 2 = AP 2 + CP 2 . Plugging in the values we have s 2 = s 2 4 + 9 16 3 R 2 3 4 s 2 = 9 16 3 R 2 s = 3 2 R The Coodinate of point C is ( - 2 a, 0 , - R/ 2) = ( - 3 2 , 0 , - 1 2 ) R . The unitvector ˆ OC = ( - 3 2 , 0 , - 1 2 ) . The coodinate of point A is ( - a,s/ 2 , - R/ 2) = ( 3 / 4 , - 3 / 4 , - 1 / 2) R . The unitvector ˆ OA = - ( 3 4 , - 3 4 , - 1 2 ) . For point B we have the coordinates as ( a,s/ 2 , - R/ 2) and the unit vector ˆ OB = ( 3 4 , 3 4 , - 1 2 ) . Let us consider the free body diagram of the Sphere. The contact forces

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## This note was uploaded on 11/06/2011 for the course MAE 101 taught by Professor Orient during the Spring '08 term at UCLA.

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MAE 101 homework 3 sol - MAE 101: Statics and Strength of...

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