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### Transcript of Tutorial 5 solution - Indian Institute of Technology … G W N 1 F 1 N 2 F 2 N 2 F 2 P N 3 F 3...

1

G

W

N1

F1

N2

F2

N2

F2 P

N3

F3

N2

F2 P

N3

F3

Consider a situation where the pipe and the wedge are in static equilibrium. Both of them have to satisfyequations of equlibrium given next.

N1 −N2 sin θ − F2 cos θ = 0 (1)

−F1 −W +N2 cos θ − F2 sin θ = 0 (2)

R(F1 − F2) = 0 (3)

F3 +N2 sin θ + F2 cos θ − P = 0 (4)

N3 −N2 cos θ + F2 sin θ = 0 (5)

Note that equation (3) is moment summation equation for the pipe about point G and R is the outer radiusof the pipe. Note that similar to last tutorial, moment equation for the wedge is not useful. Using the aboveequations, W = 50 × 9.81N and θ = 15o, we get the following relations.

N1 = 1.303F2 + 131.429 (6)

N2 = 1.303F2 + 507.803 (7)

Using F1 = F2 and µs = 0.2, the laws of dry friction can be written as given next.

N1 ≥ 5F2 (8)

N2 ≥ 5F2 (9)

It may be noted from equations (6 and 7), that inequalities (8 and 9) are satisfied when F2 = 0. Further,when F2 is increased above 0, we march towards violating the inequalities. Furthermore, from equations (6and 7), we have N2 > N1. Hence, inequality (8) get violated first. It means, in the impending motion, pipeslips across vertical wall whereas pipe will not slip on the wedge. However, pipe will roll without slip on thewedge. While it is not asked in the question, one can similarly see that in the impending motion, the wedgewill slip over the ground.At the impending motion, inequality (8) becomes just an equality. Using it in addition to equations (1–5)and F3 = 0.2N3, we can solve for seven unknowns including P . We get P = 282.972N .

θ = 15o

(4)

2

10mma

a

a = 0.0127mm∆RShaft

∆RBase

Consider the free-body diagram of a typical ball (seen from outside of the bearing) of the ball-bearing. Theweight of the ball is neglected. As per the basic theory of rolling friction, the point of action of the contactforces are offset by a. Since there are only two forces, they must have the same line of action.If the vertical load supported by the ball is ∆P , then from the geometry, the horizontal component of

the reaction is2 × 0.0127

10∆P . The torque on the shaft due to the reaction of the ball on shaft becomes

−30mm2 × 0.0127

10∆P where vertically upward direction is taken as positive. The net torque due to reac-

tions of all the balls is −30mm2 × 0.0127

10P = −190.50Nmm. The applied torque T should have a value of

190.50Nmm to balance rolling resistance.

(3)

3

(2)

The applied torque of 50(15 + 15)Ncm has to balance torque −Mscrew due to reaction at the screw and−Mthrust due to friction forces at the collar, i.e.,

50(15 + 15)Ncm = Mscrew +Mthrust (10)

Consider the situation where the impending motion is lifting up of the weight. From the theory of squarescrew, we have

Mscrew =RW tan(θ + φs)

=(2 cm)W tan

(tan−1

(1 cm

4π cm

)+ tan−1 (0.05)

)=0.26W cm (11)

From the theory of thrust bearings, at the impending motion, we have

Mthrust =2RWµ

3

=2(2 cm)W0.1

3=0.13W cm (12)

From equations (10), (11) and (12), we get W = 3811.7N .

4

(1)

Let TH and TW be the tensions of the rope on the side of the hand and on the side of the waist, respectively.Ropes are roughly vertical and vertical acceleration of the climber is negligible. Hence, force summationequilibrium equation along vertical direction gives

TH + TW = W (13)

Case (a) Minimum force required to maintain the position corresponds to the situation where the impendingmotion is loss of altitude by the climber. In such a motion, rope passes from hand-side to waist-side. Hence,

TWTH

= eµsθ = e0.2π = 1.87 (14)

From equations (13) and (14), we get TH = 0.35W .

Case (b) Minimum force to gain altitude corresponds to the situation where the impending motion is gainof altitude by the climber. In such a motion, rope passes from waist-side to hand-side. Hence,

THTW

= eµsθ = e0.2π = 1.87 (15)

From equations (13) and (15), we get TH = 0.65W .