. Force in link DF (magnitude)
Q: -2 m- 40 kN -2 m- 50 kN H I A D -2 m- C 1.5 m -30 kN 1.5 m B 40 kN 1.5 m ļ
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Q: What can the beam shear stress equation that was derived in Sec. 5.8 be used to calculate? O the…
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Q: What can the beam shear stress equation that was derived in Sec. 5.8 be used to calculate? the…
A: Given = multiple choice question
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- The figure shows a steel bar being processed by a rolling mill. Given that P=80kN and r =0.016, determine the force F required to advance the bar at a constant speed.Consider a floor crane operated by a driver which carries a drum as shown in figure. The total weight of floor crane and driver is 2700 kg acting at the point of center of gravity, G. mm; = 1 = Considering L1 = 14m, L2= 3 m L3 - 6 m, 14 = 2 m LS = 1 m; L6=7m 1. If the weight of the drum lifted by the floor crane is 800 kg, determine the total normal reaction on the rear wheels at A and total normal reaction on the front wheels at B when the boom is inclined at an angle of 25" with the horizontal, 2. Utilizing the same data given above for the same position of boom, determine the largest weight of the drum that can be lifted without causing the crane to overturn. (Hint: When the floor crane tends to overturn about point B, the wheel at A will leave the ground). Note: 1. Consider the entire system as coplanar 2. Take g 9.81 m/s wherever necessary = 9 Total normal reaction on the rear wheels at A= Total normal reaction on the front wheels at B= Largest weight of the drum=. The small piston of a hydraulic lift (g. 1) has a cross-sectional area of 3.00cm2, and its large piston has a cross-sectional area of 200cm2. What downward force of magnitude F1 must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0kN?
- The small piston of a hydraulic lift has a cross-sectional area of 3.00cm^2,and its large piston has a cross-sectional area of 200cm^2.What downward force of magnitude F1 must be applied to the small piston for the lift to raise a load whose weight is Fg= 15.0kN?A mass M is under tension from wires CD and DE. If CD is 25º from the horizontal while DE is 30º determine the safe value of mass m if the cross sectional areas of the two wires are 80 and 100mm^2 respectively. Maximum allowable stress for wire CD is 30 MPa and wired DE is 40MPa.An athlete of 60 kg and 1.70 m tall performs the ring exercise called "the Christ", in which he keeps his body immobile with his arms extended horizontally as shown in Fig.-Prob.17(a). The angle with the vertical of the cords from which the rings hang is theta = 10. (a) Determine the tension T in the cords holding the rings. (b) Considering each arm of the athlete as a rigid horizontal bar subjected to the forces indicated in Fig.-Prob.17(b), calculate the value of the components RX and RY, the value of the reaction “R” and of the beta angle.- W is the weight of the arm, applied in the middle of its length.- R is the reaction at the shoulder joint O.- RX and RY are the horizontal and vertical components respectively of the reactionapplied at the shoulder joint O.- Note: Consider that the mass of each arm of the athlete is 3% of the total mass, and thatthe length of the arm is equal to 35% of its height. Fig.-Prob.17(a): Man performing the position of "The Christ"Fig.-Prob.17(b):…
- In the figure shown, the large piston on the left has a diameter of D1 = 9.00 cm.The small piston on the right has a diameter of D2 = 4.00 cm.A lever arm pivot is located a = 5.00 cm to the left of the small piston.A small cylinder hangs from the end of the lever arm a distance b = 25.0 cm to the right of the small piston.In the absence of friction, determine the weight of the cylinder on top of the large left piston which can be supported in equilibrium with the small mass (m = 0.350 kg)hanging from the far right lever arm.Provide a force vector diagram (free body diagram).4.00 kg block is attachedto a vertical rod by meansof two strings. When the systemrotates about the axis of the rod,the strings are extended as shownin Fig. and the tensionin the upper string is 80.0 N. (a)What is the tension in the lowercord? (b) How many revolutionsper minute does the system make?(c) Find the number of revolutionsper minute at which the lowercord just goes slack. (d) Explainwhat happens if the number ofrevolutions per minute is less thanthat in part (c).A ball of mass mand radius Rrests at the edge of a step barrier of height has shown in thefigure. There is a horizontal force Tapplied to the ball from its top through a string wrapped around the ball to lift it above the barrier as shown in the figure. What is the minimum tension required to begin to lift the ball off the ground? (g = 9.8 m/s2, m = 1 kg, R = 0.5 m, h = 0.2 m).
- 2-D PROBLEM: A spherical tank with a radius of 3.41m that has a mass of 28928.644kg was supported in the system shown. Assuming friction and the bar weight is negligible, determine the following: β: 21.774° θ: 46.452° 1. The reaction @ point D. 2. Tension @cable BE 3. The x and y-component of the reaction @ A 4. The reaction @ Ad=1.50m, cable doesnt stretch and length of it between the 2 springs is 1.80m, frictionless system. m is mass of hanging weight and mp is mass of pulley.A disc (weight 50 N) is supported by two frictionless surfaces. AngleABC ¼ 90. Determine the forces between the disc and each surface usingthe graphical approach. How the result depends on the radius of the disc?