Module 3 friction 2 - Lecture notes 1 PDF

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Friction - 2...

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Mechanics Module III: Friction

Lesson 13: Friction - II The configuration of forces at a frictional contact may be visualized graphically through the concept of the friction cone and friction angle.

Figure 1:

When the friction force f = fmax = µs N (impending slippage condition), the resultant force at the point of contact must lie on the surface of a cone of semi-cone angle φs = tan−1 (fmax/N ) = tan−1 (µs ), as shown in Fig. 1. For any other static (no slip) condition, the resultant force must lie within the cone. When the surfaces are in relative sliding, the magnitude of friction force gets fixed at f = µk N . Hence, the resultant force at the contact must lie on the surface of a cone of semi-cone angle φk = tan−1 (f /N ) = tan−1 (µk ).

Figure 2:

This cone is also shown in Fig. 1. Using the concept of the friction cones, certain friction problems may be reduced to problems of geometry. Problem 1 What should be the minimum value of µs so that the spool of weight W can be pulled up the incline by pulling the string wound on the spool, as shown in Fig. 2. Solution The FBD of the spool when it just leaves the horizontal ground contact is shown in Fig. 3. The usual method for this problem is first discussed. Considering force and moment equilibrium X

Fy = 0 ⇒ N − W cos θ = 0 ⇒ N = W cos θ X

Fx = 0 ⇒ P − f − W sin θ = 0 2

Figure 3:

X

MO = 0 ⇒ r1 P − r2 f = 0

Solving the last two equations simultaneously, we have f=

W sin θ r2 −1 r1

and

P =

W sin θ 1 − rr12

In order to support this friction force f , µs N ≥ f ⇒ µs =

r1 tan θ . r2 − r1

Now we discuss the geometric method. Reconsider the FBD of the spool as shown in Fig. 4. For equilibrium, the resultant force vector F at the contact must pass through the point A which is the intersection point of the vectors W and P. At this configuration of F, the angle φ must not be greater than the semi-cone angle φs of the static friction cone, i.e., φ ≤ φs ⇒ tan φ ≤ tan φs ⇒

r1 tan θ AB ≤ µs ⇒ µs ≥ . BC r2 − r1 3

Figure 4:

Finally, the force magnitudes P and F can be determined by taking moments about C and B, respectively.

Problem 2 A thin rigid loop of mass m and radius r is suspended from a frictional support at A and loaded tangentially by a vertically downward force, as shown in Fig. 5. Determine the maximum angle β at which slipping starts. Also, determine the corresponding force P . Take the coefficient of static friction as µs . Solution The FBD of the loop at impending slippage condition is shown in Fig. 6. The net contact force at A is indicated by the vector R. At impending slippage,

4

Figure 5:

Figure 6:

5

Figure 7:

R must lie on the friction cone with semi-cone angle φs = tan−1 µs , i.e., β = φs = tan−1 µs Moment balance equation about A reads X

mgµs MA = 0 ⇒ (r sin β )mg − (r − r sin β)P = 0 ⇒ P = p 1 + µ2s − µs

Problem 3 A block of 50 kg, guided in a frictionless vertical guide, is supported by a light rod of length 300 mm due to contact friction, as shown in Fig. 7. Find the maximum value of x for which the arrangement is statically stable. Also, find the friction force acting at the contacts when x = 75 mm. Solution Since the rod AB may be considered to be massless, it is a two-force member. Hence, the contact forces at A and B are along the direction AB, as shown in the FBD in Fig. 8. At impending sliding condition at any contact, the corre6

Figure 8:

sponding force vector F must lie on the static friction cone. Since µA < µB , impending slippage will occur first at A. Hence, the maximum value of φ for static equilibrium occurs when φmax = tan−1 µA = 16.7◦. The corresponding maximum value of x is obtained from µA xmax = sin φmax = p 0.3 1 + µ2A ⇒ xmax = 0.086 m. When x = 0.075 m, sin φ = 0.075/0.3 ⇒ φ = 14.5◦. From force equilibrium, X

Fy = 0 ⇒ F cos φ − 50g = 0 ⇒ F =

50g cos φ

Now, the friction force is given by f = F sin φ = 50g tan 14.5◦ ⇒ f = 126.6 N. 7

Figure 9:

Problem 4 A rear wheel driven van has a mass of 1900 kg and identical wheels of diameter 650 mm, as shown in Fig. 9. If the coefficient of static friction between the wheels and the road surface is µs = 0.6, and the rear wheel can develop a maximum torque of 3 kNm, what is the maximum height h of the curb the van can overcome starting from rest in the configuration shown. Also, determine the wheel torque required to climb the kerb. Solution The maximum driving effort/frictional force at the rear wheel will be required to make the van just climb the curb. The FBD in this configuration is shown in Fig. 10. Note that the force F on the front wheel is normal to the wheel since it is not driven. If the contact at the rear wheel in this configuration is in impending slippage condition, the friction angle φ = φs = tan−1 µs . For equilibrium, the line of force of the vector F must pass through the point A which is the intersection of the lines of action of the vectors R and W. It 8

Figure 10:

can be observed that for A determined with φ = φs , the resulting curb height will be the highest. Now from geometry tan α = (

1.2 − rw )/1.6 = 1.047 tan φs

where rw = 0.325 m is the radius of the wheel. Hence, α = 46.3◦. Now, sin α = (rw − hmax )/rw ⇒ hmax = 0.09 m. We now check if this curb height can be overcome by the van, given that the rear wheel driving torque τ = rw ≤ 3000 Nm. Moment equilibrium about B yields X

MB = 0 ⇒ (1.2)1900g − (0.325 + 1.25 × 2.8)f = 0 ⇒ f = 6868.6 N.

Hence, the required rear wheel torque τ = rw f = 2232.3 Nm. This torque can be achieved by the van.

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