346_PartUniversity Physics Solution

# 346_PartUniversity Physics Solution - Equilibrium and...

This preview shows pages 1–3. Sign up to view the full content.

Equilibrium and Elasticity 11-5 11.13. IDENTIFY: Apply the first and second conditions of equilibrium to the strut. (a) SET UP: The free-body diagram for the strut is given in Figure 11.13a. Take the origin of coordinates at the hinge (point A) and y + upward. Let h F and v F be the horizontal and vertical components of the force F ! exerted on the strut by the pivot. The tension in the vertical cable is the weight w of the suspended object. The weight w of the strut can be taken to act at the center of the strut. Let L be the length of the strut. EXECUTE: y y Fm a = v 0 Fww −= v 2 Fw = Figure 11.13a Sum torques about point A. The pivot force has zero moment arm for this axis and so doesn&t enter into the torque equation. 0 A τ = () sin30.0 ( /2)cos30.0 ( cos30.0 ) 0 TL w L w L °− ° − ° = sin30.0 (3 /2)cos30.0 0 Tw °= 3 cos30.0 2.60 2sin30.0 w ° == ° Then x x a = implies h 0 TF and h 2.60 . = We now have the components of F ! so can find its magnitude and direction (Figure 11.13b) 22 hv FF F =+ (2.60 ) (2.00 ) 3.28 = v h 2.00 tan 2.60 θ 37.6 = ° Figure 11.13b (b) SET UP: The free-body diagram for the strut is given in Figure 11.13c. Figure 11.13c The tension T has been replaced by its x and y components. The torque due to T equals the sum of the torques of its components, and the latter are easier to calculate.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
11-6 Chapter 11 EXECUTE: 0 ( cos30.0 )( sin45.0 ) ( sin30.0 )( cos45.0 ) A TL τ =+ ° °− ° (( / 2)cos45.0 ) ( cos45.0 ) 0 wL °= The length L divides out of the equation. The equation can also be simplified by noting that sin45.0 cos45.0 . ° Then (cos30.0 sin30.0 ) 3 /2. Tw ° = 3 4.10 2(cos30.0 sin30.0 ) w == ° x x Fm a = h cos30.0 0 FT −° = h cos30.0 (4.10 )(cos30.0 ) 3.55 w w = ° = y y a = v sin30.0 0 FwwT −−− ° = v 2 (4.10 )sin30.0 4.05 Fw w w ° = From Figure 11.13d, 22 hv FF F (3.55 ) (4.05 ) 5.39 Fww w = v h 4.05 tan 3.55 θ 48.8 = ° Figure 11.13d EVALUATE: In each case the force exerted by the pivot does not act along the strut. Consider the net torque about the upper end of the strut. If the pivot force acted along the strut, it would have zero torque about this point. The two forces acting at this point also have zero torque and there would be one nonzero torque, due to the weight of the strut. The net torque about this point would then not be zero, violating the second condition of equilibrium.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

346_PartUniversity Physics Solution - Equilibrium and...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online