Keeping a broken leg bone straight while it is healing sometimes requires

traction, which is the procedure in which the bone is held under stretching

tension forces at both ends to keep it aligned. Consider a leg under

tractional tension as shown in ▼ Figure 4.16. The cord is attached to

a suspended mass of 5.0 kg and runs over a pulley. The attached cord

above the pulley makes an angle of θ = 40° with the vertical. Neglecting

the mass of the lower leg and the pulley and assuming all the strings are

ideal, determine the magnitude of the tension in the horizontal cord. 

THINKING IT THROUGH. The pulley is in a static equilibrium

and thus has no net force on it. If the forces are summed both vertically

and horizontally, they independently should add to zero.

This should allow the tension in the horizontal string to be found.

9.An 80-kg person stands on one foot with the heel elevated

(▶ Figure 4.17a). This gives rise to a tibia force F1 and an Achilles

tendon “pull” force F2 as illustrated in Figure 4.17b. Typical angles

are θ1 = 15° and θ2 = 21°, respectively. (a) Find general equations for

F1 and F2, and show that θ2 must be greater than θ1 to prevent damage

to the Achilles tendon. (b) Compare the force applied by the

Achilles tendon with the weight of the person.

THINKING IT THROUGH. This is a case of static translational

equilibrium, so the x- and y-components can be summed to get

equations for F1 and F2.

Transcribed Image Text:

mg (weight) 71 (pulley) 1