Name. MECH 223 Engineering Statics. Midterm 1, February 24 th 2015

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1 1 Name MECH 223 Engineering Statics Midterm 1, February 24 th 2015 Question 1 ( points) (a) (5 points) Form the vector products B C and B C (where B = B ) and use the result to prove the identity sin α cos β = 1 2 sin(α + β) + 1 sin(α β) 2 y α C β β B x B By definition the cross product of B C is B C sin θ where θ is the angle between the two vectors when they are joined tail to tail and the direction of the resulting vector is perpendicular to the plane containing the two original vectors. In the case of the figure in this problem the angle between B and C is θ=α-β B C = B C sin(α β). The angle between B and C is θ=α+β B C = B C sin(α + β). The direction of both vectors is out of the page. On the other hand we can do a cross product by Cartesian components of each vector. B = B (cos β i + sin β j ), B = B (cos β i sin β j ) and C = C (cos α i + sin α j ). B C = B C (cos β sin α k sin β cos α k ) (product of i i and j j is equal to 0). B C = B C (cos β sin α k + sin β cos α k ) Adding the two cross products and equating the two results produces: B C sin(α β) + B C sin(α + β) = B C (cos β sin α sin β cos α) + B C (cos β sin α + sin β cos α) cos(β) sin(α) = 1 (sin(α β) + sin(α + β)) QED 2

2 2 (b) (5 points) Define the dot product of two vectors. Using this definition, prove that A B = A x B x + A y B y + A z B z. Hint: first, define the dot product of the Cartesian unit vectors: i i, i j, By definition the dot product of A B is A B cos θ where θ is the angle between the two vectors when they are joined tail to tail. i i = 1; j j = 1; k k = 1 i j = 0; i k = 0; j k = 0 If A = A x i + A y j + A z k and B = B x i + B y j + B z k then: A B = A x B x + A y B y + A z B z (c) (5 points) Using the appropriate sketch for II, prove that I is equivalent to III. Provide a formula for M. Define any necessary auxiliary quantity on the sketch.

3 3 (d) (5 points) Using relevant sketches, explain the difference between free, sliding, and fixed vectors in mechanics. Give examples of physical quantities represented by these vectors. Free vector: point of application and line of action are not important (Example: velocity of body moving without rotation) Sliding Vector: line of action is important (Example: force in rigid body mechanics). Fixed vector: point of application is important (Example: force in elastic bodies) (e) (Bonus, 5 points) What is the physical meaning of the vector product of two vectors? What is the physical meaning of the triple product of three vectors? Use sketches to illustrate your explanation if necessary. Vector product a vector which magnitude corresponds to the area of the parallelogram created by the two original vectors (alternatively, sum of projections of moment on the three axes). Triple product volume of the parallelepiped created by the three vectors.

4 4 Question 2 ( points). A winch puller AB is used to strengthen a fence post. Knowing that the tension in cable BC is 260 lb, legth a is 8 in., length b is 35 in., and length d is 76 in., determine the moment about point D of the force exerted by the cable at C by resolving that force into horizontal and vertical components applied (a) (10 points) at point C, (b) (10 points) at point E. Note: Solutions by other methods will not carry any credit.

5 5 (c) (Bonus, 5 points) Using the scalar product, what is the distance between D and the cable? DF perpendicular to CE, and DF = d DF CE = x F ( 76 8) + y F ( 35) = 0 y F = 84 x 35 F On the other hand, we can get a formula for EC: y = (x + 76) x F = x F + 84 E F (x F,y F) d y D C x x F = 11.2 and y F = 27.0 d = x 2 F + y 2 F = 29.2 Question 3 (30 points). Four horizontal forces are applied as 200 N A shown to a vertical cast iron arm. You can neglect the width of the base. (a) (10 points) Determine the resultant of the forces and the distance from the ground to its line of action when a = 150 mm and the force Q = 0 lb. b Q B C D 600 N 400 N a 150 mm 300 mm R = F = ( )N i = 800 N i To determine the line of action, first convert the system of forces to an equivalent force + couple at D, then shift the line of action to eliminate the couple. Line of action is applied at height y (up) from D. M = Ry = 600N 0.3m + 400N 0.15m 200N 0.45m = 150 N m

6 6 y = m = mm (b) (10 points) What is the distance a when the force Q = 0 lb if it is known that the distance between the line of action of the resultant and the ground is 250 mm. This time we know y. The moment created by the resultant around D needs to be equal to the sum of the moments of all the forces around D. M = Ry = 800N 0.25m = 200N m = 600N 0.3m + 400N a 200N 0.45m ( )N m a = = 0.275m = 275 mm 400 N (c) (10 points) If the force Q = 500 lb and a = 150 mm, what is the distance b and the resultant force, if it is known that the line of action of the resultant of the forces passes at the base of the arm. Since the line of action passes through the base of the arm, it means that the equivalent system of forces at D doesn t have a couple. Similar to (a): R = F = ( )N i = 300 N i However, now: M = 600N 0.3m + 400N 0.15m 200N 0.45m 500N b = 0 ( )N m b = = 0.3m = 300mm 500N

7 7 Question 4 (30 points) (a) (15 points) Determine the unstretched length of spring AC (8 points), the force exerted by it (3 points), and the tension in the cord AB (4 points) if a force P = 80 lb causes the angle θ = 60 for equilibrium. Cord AB is 2 ft long. Take k = 50 lb/ft. l = cos 60 = 12 From Sine law: φ 12 sin 60 = 2 sin φ φ = 30 + F y = 0; T sin 60 + F s sin = 0 + F x = 0; T cos 60 + F s cos 30 = 0 T = F s cos 30 cos 60 cos 30 sin 60 F s ( + sin 30 ) = 80; F cos 60 s = 40 lb T = 69.3 lb F s = kx 40 = 50( 12 l ) l = 2.66 ft

8 8 (b) (15 points) If the maximum force that the spring can exert before undergoing irreversible deformation is 60 lb, what is the maximum allowed force P (7 points), the corresponding tension in the cord AB (4 points), and the angle θ (4 points). F s = kx x = 60 lb 50 lb ft = 1.2 ft l = l + x = 3.86 ft l = cos θ cos θ = From Sine law: 3.86 sin 71.4 = 2 sin φ = ; θ = 71.4 φ = F y = 0; T sin θ + F s sin φ P = 0 θ φ + F x = 0; T cos θ + F s cos φ = 0 θ φ cos φ T = F s = lb cos θ P = T sin θ + F s sin φ = lb P