Rotary Homework #7

Rotary Homework #7: 1. Calculate FA and FB for the beam shown in the figure below. The downward forces represent the weights of machinery on the beam. Assume the beam is uniform and has a mass of 250 kg.

2. Three children are trying to balance on a seesaw, which consists of a fulcrum rock, acting as a pivot at the center, and a very light board 3.6 m long (figure below). Two playmates are already on either end. Boy A has a mass of 50 kg, and girl B a mass of 35 kg. Where should girl C, whose mass is 25 kg, place herself so as to balance the seesaw?

3. A 172-cm-tall person lies on a light (massless) board which is supported by two scales, one under the top of her head and one beneath the bottom of her feet (figure below). The two scales read, respectively, 35.1 and 31.6 kg. What distance is the center of gravity of this person from the bottom of her feet?

4. A 20.0-kg floodlight in a park is supported at the end of a horizontal beam of negligible mass that is hinged to a pole, as shown in the figure below. A cable at an angle of 30.0° with the beam helps to support the light. Find (a) the tension in the cable and (b) the horizontal and vertical forces exerted on the beam by the pole. 5. A shop sign weighing 245 N is supported by a uniform 155-N beam as shown in the figure below. Find the tension in the guy wire and the horizontal and vertical forces exerted by the hinge on the beam.

6. A 500-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long, uniform, 100-N rod as indicated in the figure at right. The left end of the rod is supported by a hinge, and the right end is supported by a thin cable making a 30.0° angle with the vertical. (a) Find the tension T in the cable. (b) Find the horizontal and vertical components of force exerted on the left end of the rod by the hinge.

7. A hungry 700-N bear walks out on a beam in an attempt to retrieve some “goodies” hanging at the end (Fig. P8.20). The beam is uniform, weighs 200 N, and is 6.00 m long; the goodies weigh 80.0 N. (a) Draw a free-body diagram of the beam. (b) When the bear is at x = 1.00 m, find the tension in the wire and the components of the reaction force at the hinge. (c) If the wire can withstand a maximum tension of 900 N, what is the maximum distance the bear can walk before the wire breaks?

8. {Bonus Question} What minimum horizontal force F is needed to pull a wheel of radius R and mass M over a step of height h as shown in the figure below R  h? (a) Assume the force is applied at the top edge as shown. (b) Assume the force is applied instead at the wheel’s center. Solutions: 1. The beam is in equilibrium, and so both the net torque and net force on it must be zero. From the free-body diagram, calculate the net torque about the center of the left support, with counterclockwise torques as positive. Calculate the net force, with upward as positive. Use

those two equations to find FA and FB .    F1 F2 F3   F A x x x x FB 1 2 3 4  x5 mg

t =FB( x 1 + x 2 + x 3 + x 4) - F 1 x 1 - F 2( x 1 + x 2) - F 3( x 1 + x 2 + x 3) - mgx 5

F1 x 1+ F 2( x 1 + x 2) + F 3( x 1 + x 2 + x 3) + mgx 5 FB = ( x1+ x 2 + x 3 + x 4 ) (4300 N)( 2.0 m) +( 3100 N)( 6.0 m) +( 2200 N)( 9.0 m) + ( 250 kg)( 9.8m s2 )( 5.0 m) = 10.0 m = 5925 N淮 5.9 103 N

F= FA + F B - F 1 - F 2 - F 3 - mg = 0 2淮 3 FA= F 1 + F 2 + F 3 + mg - F B =9600 N +( 250 kg)( 9.8 m s) - 5925 N = 6125 N 6.1 10 N 2. From the free-body diagram, the conditions of equilibrium are used to find the location of the

girl (mass mC ). The 50-kg boy is represented by mA , and the 35-kg girl by mB . Calculate torques about the center of the see-saw, and take counterclockwise torques to be positive.  The upward force of the fulcrum on the see-saw (F) causes no torque about the center.

1 1 t =mA g( 2 L) - m C gx - m B g( 2 L) = 0 L m- m 50 kg - 35 kg ( A B ) 1( ) 1 x x=( 2 L) = 2 (3.6 m) = 1.1 m mC 25 kg  m g    A F mCg mBg

3. The person is in equilibrium, and so both the net torque and net force must be zero. From the free-body diagram, calculate the net torque  L - x x  about the center of gravity, with counterclockwise torques as FA FB positive. Use that calculation to find the location of the center of  gravity, a distance x from the feet. mg

t =FB x - F A ( L - x) = 0 F m g m 35.1 kg x=A L = A L = A L =(1.72 m) = 9.05 10-1 m FA+ F B m A g + m B g m A + m B 31.6 kg + 35.1 kg The center of gravity is about 90.5 cm from the feet. 4. (a) Consider the torques about an axis perpendicular to the page and through the left end of the horizontal beam.

St = +(Tsin 30.0�) d = ( 196 N) d 0 ,

giving T = 392 N

(b) From SFx = 0 , H- T cos 30.0� 0 , or

H = (392 N) cos 30.0� 339 N to the right

From SFy = 0 , V+ T sin 30.0� = 196 N 0

or V =196 N - ( 392 N) sin 30.0� 0

 5. The beam is in equilibrium. Use the conditions of equilibrium to calculate  FT the tension in the wire and the forces at the hinge. Calculate torques FH about the hinge, and take counterclockwise torques to be positive. q t=Fsin q l - m g l 2 - m gl = 0   ( T) 2 1 1 2 1 m g l1 2 1 m2g 1 m gl+ m gl 1 (155 N)( 1.70 m) + ( 245 N)( 1.70 m) F =2 1 1 2 1 = 2 l T o 2 l2 sinq (1.35 m)( sin 35.0 ) l1 = 708.0 N淮 7.08 102 N

o 2 Fx= FH x - F Tcosq = 0 � F H x = F T cos q =( 708 N) cos35.0 淮 579.99 N 5.80 10 N

Fy= FH y + F Tsinq - m 1 g - m 2 g = 0

o FHy = m 1 g + m 2 g - F T sinq = 155 N + 245 N -( 708 N) sin 35.0 = - 6.092 N� 6 N( down) 6. Consider the torques about an axis perpendicular to the page through the left end of the rod.

(100 N)( 3.00 m) + ( 500 N)( 4.00 m ) St = 0 � T (6.00 m) cos 30.0 T = 443 N

SFx = 0 � R x � T sin 30.0( 443 N) sin 30.0

Rx = 221 N toward the right

SFy =0 � R y � T cos 30.0 -= 100 N 500 N 0

Ry =600 N - ( 443 N) cos 30.0� 217 N upward

7. (a) See the diagram below: (b) If x = 1.00 m, then

St )left end =0 � -( 700 N)( 1.00 m) - ( 200 N)( 3.00 m ) ( 80.0 N)( 6.00 m) + (T sin 60.0�)( 6.00 m) 0

giving T = 343 N

Then, SFx = 0 � H � T cos 60.0 0 , or H = (343 N) cos60.0� 171 N

and SFy = 0 � V � 980 N+( 343 N) sin 60.0 0 , or V = 683 N

(c) When the wire is on the verge of breaking, T = 900 N and

St )left end = -( 700 N) xmax -( 200 N)( 3.00 m ) - ( 80.0 N )(6.00 m) + 臌轾( 900 N) sin 60.0�( 6.00 m) 0

which gives x max = 5.14 m

8. (a) If the wheel is just lifted off the lowest level, then the only  forces on the wheel are the horizontal pull, its weight, and the F  2R- h contact force FN at the corner. Take torques about the corner  point, for the wheel just barely off the ground, being held in  FN Mg equilibrium. The contact force at the corner exerts no torque and so does not enter the calculation. The pulling force has a lever arm of R+ R - h =2 R - h , and gravity has a lever arm of x x , found from the triangle shown. x= R2 -( R - h)2 = h(2 R - h) R- h R t =Mgx - F(2 R - h) = 0 x Mgxh(2 R- h) h F= = Mg = Mg 2R- h 2 R - h 2 R - h

. (b) The only difference is that now the pulling force has a lever arm  of R- h . F  t =Mgx - F( R - h) = 0 R- h Mg

Mgx h(2 R- h) F= = Mg R- h R - h x