Published on 26 Jan 2015

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ENGG 202 Tutorial 4 Jan 29/30, 2013

*question is from an old ENGG 205 exam

Review – 3D vectors

1) To stabilize a tree partially uprooted in a storm, cables AB and AC

are attached to the upper trunk of the tree and then are fastened to

steel rods anchored in the ground. Knowing that the tension in cable

AB is 4.2 kN, determine (a) the components of the force exerted by

this cable on the tree and (b) the angles θx, θy, θz for this force.

ANS: F=2.46i-2.70j+2.07k (kN), (b) 54.1o, 130.0o, 60.5o

Solution Strategy: a) to find the components, determine the unit

vector for the direction AB (find position vector from the dimensions

and divide by the magnitude) then multiply the unit vector by the

force magnitude.

b) the components of the unit vector for direction AB found in the

intermediate step in (a) are the direction cosines. Therefore, to find

the angle, take the inverse cosine of each component of the unit vector. Or use the equations from

page 10 of your notes.

2D Equilibrium

2*) Given the tension force T = 10 kN, determine the

weight of the block, W, necessary for the pulley-cable

system to be in equilibrium. Neglect the weight of the

pulleys.

ANS: W = 40 kN

Solution Strategy: there are 2 cables in this system, cable 1 over pulleys C

and B and cable 2 over pulley D. each cable has a different tension.

Drawing FBDs of pulley C and pulley B and applying your equations of

equilibrium will allow you to solve for the force in cable 2 in terms of the

cable 1 tension and then to solve for the weight W.

3*) For each case (a) and (b), determine the force P required to

maintain equilibrium. The block weighs 100 N. Which cable-pulley

system is more efficient?

ANS: (a) P = 25 N, (b) P = 11.1 N (more efficient)

Solution Strategy: in case (a) there is only one cable. So drawing

the FBD for the bottom 2 pulleys together (cut through the cable 4

times and include the weight) will allow you to solve for P

For case (b) there are two cables, one that passes over the two pulleys

on the right and one that passes over the two left pulleys. Drawing

the FBDs for the largest pulley will allow you to solve for the force in

one of the cables in terms of the weight, then drawing the FBD for the

next largest pulley will allow you to solve for the force in the other

cable which is P.