Solutions 1P3200 - Lecture notes 1 PDF

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course notes for material science, including ch2 and chapter 3...

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Phys 3200 Fall 2015 Exam #1 Name:________________________________________ You may use your non-programmable calculator. You may not consult books or notes. You must show your work. Box or circle your answers. No partial credit is given unless so indicated. 1. (30 points) Knowing that = 50o and that boom AC exerts on pin C a force directed along line AC, determine the force P such that the tension in the cable is 300 lb.

Solution: In the free-body diagram for the pin at C, there are three forces, the sum of which should vanish. The three forces form a triangle with two of the angles of the triangle identified as 100o and 30o. The third is then 50o. The Law of Sines can now be used to write

P FAC 300 lb   o o sin100 sin 50 sin 30o P  386 lb

2. (25 points, Partial Credit) To keep a door closed, a wooden stick is wedged between the floor and the doorknob. The stick exerts at B a 175-N force directed along line AB. Replace that force with an equivalent force-couple system at C.

Solution: Since a stick cannot pull the doorknob, the 175 N force is in the direction from A to B, and can be written as        73i  990 j  494 k F  175   (11.5i  156 j  78.0k ) N 732  990 2  494 2

The displacement from C to B is     rB /C  (683i  860 j  0 k ) mm

The moment about C is    M C  rB / C  F       (683i  860 j ) (11.5i  156 j  78.0k ) mm  N     (67.1i  53.3 j  116k ) m  N

3. (20 points, Partial Credit) Two 150-mm-diameter pulleys are mounted on line shaft AD. The belts at B and C lie in vertical planes parallel to the yz plane. Replace the belt forces shown with an equivalent force-couple system at A.

Solution: The two forces on belt at C combine to a total force of    FC  370 N  ( sin10o j  cos10o k )    ( 64.2 j  364 k ) N and a moment about C of   M C  75  60 i mm  N . When tranferred to point A, the moment of these two forces is     o o M AC  4500i  405  370(cos10 j  sin10 k ) mm  N     (4.50i  147.6 j  26.0k ) m  N The two forces on belt at B combine to a total of      FB  [( 240cos 20o  145) j 240sin 20 o k ] N  (370 j  82.1k ) N and a moment about B of    M B  75  95i mm  N  7.12i m  N . When tranferred to point A, the  couple of these two forces is    M AB  (7.12 i 14.78 j  66.7 k ) m  N The total force at point A is      FA  FB  FC  ( 434 j  282k )N and the total couple at A is           M A  (4.50i  147.6 j  26.0k  7.12i  14.78 j  66.7k ) m  N  (11.62i 132.8 j  92.7k ) m  N

4. (25 points) A 50-kg crate is attached to the trolley-beam system shown. Knowing that a = 1.2 m, determine (a) the tension in cable CD, (b) the reaction at B.

Solution: Three forces acting on the massless steel beam sum to zero force and zero moment. The moment about B vanishes, from which we write MB  0  T 1.8sin 55 o  50*9.8 1.2 ( m  N ) T  399 N

0  BX  399N sin 55o

BX  327 N

0  By  50*9.8  399cos 55o

Or,

B  327  261  419 N 2

2

B y  261N at an angle of 180o  tan 1 (261/ 327)  141.4 o

5. (10 points, Extra Credit) A 450-lb load hangs from the corner C of a rigid piece of pipe ABCD which has been bent as shown. The pipe is supported by the ball-and-socket joints A and D, which are fastened, resectively, to the floor and to a vertical wall anc by a cable attached at the midpoint E of the portion BC of the pipe and at a point G on the wall. Determine (a) where G should be located if the tension in the cable is to be minimum, (b) the corresponding minimum value of the tension.

Solution: The total moment about the line AD, which has a direction unit vector of     12i  12 j  6 k 2  2  1   i  j  k , should vanish. AD  3 3 122  122  62 3 The moment due to the 450 lb force is     2  2  1     AD  rC / A  (450 j )   i  j  k   (12 i  12 j )  ( 450 j ) ft  lb  1800 ft  lb , which should be 3 3  3 balanced by the moment due to tension. Maximum moment is applied at point E, if the direction of the force (i.e. EG) is made parallel to    2  2  1      2 1  2  (6i  12 j )  i  j  k    4i  2 j  4k  6 ft   i  j  k  3 3  3 3  3  3 Since 6 ft is the lever arm, the minimum value of tension is

1800 ft  lb / 6 ft  300lb  To find the location G, we notice that the z-component of rG / E is -6 ft, which means that     rG/ E  6.00 i  3.00 j  6.00 k ft

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or    rG/ D   12.0 i  3.00 j  ft

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