Question

Q2.

Context Rich Problem. Show your work and reasoning.

Torque:

The automatic flag raising system on a horizontal flagpole attached to the vertical outside wall of a tall building has become stuck. The management of the building wants to send a person crawling out along the flagpole to fix the problem. Because of your physics knowledge, you have been asked to consult with a group to decide whether or not this is possible. You are all too aware that no one could survive the 250 foot fall from the flagpole to the ground. The flagpole is a 120 lb steel I-beam which is very strong and rigid. One side of the flagpole is attached to the wall of the building by a hinge so that it can rotate vertically. Nine feet away, the other end of the flagpole is attached to a strong, lightweight cable. The cable goes up from the flagpole at an angle of 30° until it reaches the building where it is bolted to the wall. The mechanic who will climb out on the flagpole weighs 150 lbs including equipment. From the specifications of the building construction, both the bolt attaching the cable to the building and the hinge have been tested to hold a force of 500 lbs. Your boss has decided that the

worst case is when the mechanic is at the far end of the flagpole, nine feet from the building.

Draw your diagram below.

Answer 1

1. Draw the free-body diagram of the steel bar. Indicate in your drawing the axis of rotation. Possible pivot/axis to evaluate torque Fhinge = 300 Fw,steel Fw,man 2. From your diagram, enumerate three dynamical equations that will satisfy static equilibrium. 1st Condition: F = Fninger - T cos 0 = 0 EF, = Fringey +Tsino – Fristeet – Exman = 0 2nd Condition: (about the end of the pole)

Ei = 'Fusteet – LFningey = 0 (3) 2 3. Wisely plan a solution based on the above equations. (Again, it is wiser to find a final working equation before doing any substitution of the given values.) Target Quantities: Fhinge and T (since Fbolt = T) From (3), we may solve for the y-component of Fhinge: -F steel = Lr hingey w.steen Fringes = 3m.co Substitution of (4) to (2): Em steet8 +T sin 0 – msteel 8 - mmang = 0) --meeg +Tsin NI -mwang=0 T sin = m g + steel 2 So we can solve T directly from (5), m mang+ m steel T=- sin e Note that Fringe = V(Fningex]* + (Fninger) To solve for the x-component Fhinge, substitute (6) to (1): Fringex = T coso

Fran m mang + .us + 5Mad & costo m man + M steel Fhingex = m steel - cos =- sin (8) tan 0 8 Then (4) and (8) to (7): Iman 171 +2M steel 2 Fhinge 9 + -msteel 8 12 tan 4. Execute your strategy. Is the force on the cable and the hinge equal or less than the specified 500lbs x g? Check: mman = 150 lbs. msteel = 120 lbs. 0-300 Then, (150lbs)g +- (120lbs )g - = (420lbs)g T = sin 30° 150lbs + +1201bs & 1 + ((120165)g hinge tan 30° 50lbs += 120lbs g 2 =g + 3 12015) = (368/h) + - 120lbs = (368lbs g tan 30° 2 YES! Both the tension on the cable/force on the bolt and the force on the hinge are below the limit specifications.

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