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Statics of Rigid Bodies | Friction | Moment of Inertia
Typology: Exercises
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Problem 1. Determine the moments of inertia of the rectangular area shown about the centroidal x 0 - and y 0 -axes, the centroidal polar axis through C. Solution: For the moment of inertia about the x 0 - axis, a horizontal strip of area bdy can be used so that the y-coordinate is constant throughout b: Following the same process, but this time about other axis, the moment of inertia about the y 0 - axis is:
For the centroidal polar moment of inertia about C. (Answer) Problem 2. Calculate the moments of inertia of the area of a circle about a diametral axis and about the polar axis through the center. Specify the radii of gyration. Solution: A differential element of area in the form of a circular ring may be used for the calculation of the moment of inertia about the polar axis through O since all elements of the ring are equidistant from O. The elemental area is and thus,
Solution: The solution for the problem is very similar to that of a composite section, only in this case, some parts have to be deducted from a larger regular shape. There are three subareas presented in the figure – a rectangle, a quarter circle and a triangle. The latter two subareas are to be subtracted from the first as shown next. The calculation is presented in the table shown:
Using the Transfer Formula, the Moments of Inertia can be determined:
Problem 2. It is observed that when the bed of the dump truck is raised to an angle of θ = 25° the vending machines on it will begin to slide off the bed. Determine the static coefficient of friction between the vending machine and the surface of the truckbed. Solution. The FBD of the vending machine is shown including the given dimensions. We will assume its weight to be W. The dimension x is used to locate the position of the resultant normal force N. There are four unknowns, N, F, μs, and x. Equations of Equilibrium. ∑Fx = 0; W sin 25° - F = 0 (1) ∑Fy = 0; N - W cos 25° = 0 (2) ∑MO =0; -Wsin25°(2.5)+Wcos25°(x)=0 (3) Since slipping impends at u = 25°, using Eqs. 1 and 2, we have Fs =μsN;Wsin25°=μs(Wcos25°) μs = tan 25° = 0.466 Ans The angle of θ = 25° is referred to as the angle of repose, and by comparison, it is equal to the angle of static friction, θ = φs. Notice from the calculation that u is independent of the weight of the vending machine, and so knowing u provides a convenient method for determining the coefficient of static friction. Note: From Eq. 3, we find x = 1.17 ft. Since 1.17 ft < 1.5 ft (location of center of gravity), indeed the vending machine will slip before it can tip.
Problem 3. A uniform 10-kg ladder rests against a smooth wall at point B, and the end A rests on the rough horizontal plane for which the coefficient of static friction is μs = 0.30. Determine the angle of inclination θ of the ladder and the normal reaction at B if the ladder is on the verge of slipping. Solution. As shown on the FBD, the frictional force FA must act to the right since impending motion at A is to the left. Equations of Equilibrium and Friction. Since the ladder is on the verge of slipping, then FA = μsNA = 0.3NA. By inspection, NA can be obtained directly. ∑Fy=0; NA -10(9.81)N=0NA =98.1N Using this result, FA = 0.3(98.1 N) = 29.43 N. Now NB can be found. ∑Fx=0; 29.43N-NB = NB =29.43N=29.4N Ans. Finally, the angle θ can be determined by summing moments about point A. ∑MA =0; (29.43N)(4m)sinθ-10(9.81)Ncosθ=0 sin θ / cos θ = tan θ = 1. θ = 59.04° = 59.0° Ans.
Problem 5. The turnbuckle shown below has a square thread with a mean radius of 5 mm and a lead of 2 mm. If μs = 0.25, between the screw and the turnbuckle, determine the moment M that must be applied to draw the end screws closer together. Solution. Since friction at two screws must be overcome, this requires M = 2[rW tan(θ + φs)] (1) Here W = 2000 N, φs = tan
Problem 6. The maximum tension that can be developed in the cord shown is 500 N. If the pulley at A is free to rotate and the coefficient of static friction at the fixed drums B and C is μs = 0.25, determine the largest mass of the cylinder that can be lifted by the cord. Solution. Lifting the cylinder, which has a weight W = mg, causes the cord to move counterclockwise over the drums at B and C; hence, the maximum tension T 2 in the cord occurs at D. Thus, F = T 2 = 500 N. A section of the cord passing over the drum at B as shown. Since 180° = π rad the angle of contact between the drum and the cord is β = (135°/180°)π = 3π/4 rad. We have T 2 = T 1 e μsβ ; 500 N = T 1 e 0.25[(3>4)π] Hence, T 1 =500N/e 0.25[(3>4)π] =500N/1.80=277.4N Since the pulley at A is free to rotate, equilibrium requires that the tension in the cord remains the same on both sides of the pulley. The section of the cord passing over the drum at C is shown. The weight W < 277.4 N. Why? We obtain T 2 = T 1 e μsβ ; 277.4 N = We 0.25[(3>4)π] W = 153.9 N so that m=W/g=153.9N/9.81m/s 2 =15.7kg Ans.
