2 blocks with mass, 2 massless pulleys, give equation of motion for x(t): x(t) m, m, y(1)
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- Problem 1 A fireman sits on the top step of a ladder, which is assumed to be massless. He can be considered as a point mass m on the end of the ladder. The ladder is rotated up by a driving torque at the point x = 0 and simultaneously extended with the constant velocity v,ej. The ladder has the length r(t) at time t. The motion occurs under the influence of gravity. Given: m; vret = const. ; p ; ở ; ö ; r; g Find: 1. By using the polar coordinates system, determine the inertial forces (d'Alembert's forces) acting on the fireman 2. Determine the internal forces (L(x), Q(x), M(x)) in the ladder during the motion as a function of , o, ở and x. m. Vrel r(t) Driving TorqueGiven that the slot (for the cord) in the cylinder with 17.6 kg in the figure below has a negligible effect on Ic, find the time required for C to move 3.7 m down the incline if 0 = 55°. R = 1.4 m,r = 0.6 m, and u = 0.11. Cord R Figure is from "Engineering Mechanic An Introduction to Dynamics", McGill and King.3. Quarter car suspension model The figure below a model of the suspension of a car. r(t)is the position of the mass of the car from its equilibrium position while z is the profile of the road. Note that z is a function of the distance y along the road, but you need ż(t) which you can obtain from z(y) using the chain rule. The car is traveling to the right on the road at a speed v. Assume the mass of the car is 250 kg, the spring constant is 16 kN/m, and damping constant is 1000 Ns/m. (a) Derive the differential equation for the position of the car r(t). (b) If z(t) = Zoejot, determine the amplitude and the phase of the particular solution. (c) (MATLAB) If the car is travelling with constant velocity along a road with rolling hills that can be represented as a cosine function with amplitude Zo and wavelength A (wavelength is the distance between adjacent crests or adjacent troughs, determine the vertical motion of the car as a function of time. X(t) Z m 1 C Figure 1: Quarter car…
- 5. There are two masses with unequal weight. The masses are tied by a rope and they are connected over two pulleys as shown in Figure 1. When the system is released, heavier mass accelerates downward and lighter accelerates upward. Assume frictionless pulleys, find the theoretical tension of the thread and the theoretical acceleration. Masses are M1= 6.00 kg and M2= 2.00 kg. Two discs are allowed to move on inclined air table with dimensions a= 11cm, b=17cm, c=60cm. (Hint: You can use free body diagram and M1 b Newton' W.B. M2 s second law of a motion equation (F=m.a) to solve tension, g=980cm/s?.)5. The principles of Statics also apply to objects in uniform rectilinear motion (ả = 0). A box with weight W slides at a constant speed down an incline of 30° (the angle of repose). If a horizontal force P is applied, the box can be pushed up the ramp at constant speed. What is P? V3w ans:The following mass-and-spring system has stiffness matrix K. The system is set in motion from rest (x, '(0) = x2'(0) = 0) in its equilibrium position (x, (0) = x2(0) = 0) with the given external forces F, (t) = 0 and F, (t) = 270 cos 4t acting on the masses m, and m,, respectively. Find the resulting motion of the system and describe it as a %3D superposition of oscillations at three different frequencies. k2 mi ww m2 k3 - (k, + k2) k2 K= k2 - (k2 + k3) m, = 1, m, = 2; k, = 1, k, = 6, k3 = 2 %3D Find the resulting motion of the system. X4 (t) = X2(t) (Type exact answers, using radicals as needed.)
- 4. The figure below shows a bulldozer. Notice that the track on the bulldozer (the tread that wraps around the wheels) is kinematically similar to the conveyor belt in the previous problem. D В E Suppose the person operating the bulldozer knows the distance between the two main wheels on the side of the vehicle. She looks out the window and, with the aid of a stopwatch, notices that the track is moving forward from her perspective at a constant speed v. Furthermore, as the bulldozer drives forward, the track does not slide or skid on the ground. In answering the sub-questions below, use the idea contained in Equation (2) and express answers in terms of basis (ê1, ê2). You may utilize results from Problem 3. (a) Using the terminology in (2), what does the quantity v directly represent? (b) Let C be a point on the track in contact with the ground. What is the velocity of point C as observed by a stationary person on the ground watching the bulldozer? Why? (c) Find the velocity of point…0.5 m, k = 500 N/m, and k = 40 N.m/rad, In the system shown, m = 10 kg, r = 0.3 m, Ko where ko is the radius of gyration of the pulley about point 0. There is no slippage between the cord and the pulley. Replace the system with an equivalent { a) torsional spring and mass moment of inertia at point 0, and b) translational mass and spring at point A. т, Ко kt А k m wwQ2) Consider a mass (m=5 kg) connected to a massless rod of length (L=2 m) to swing about point O, as shown in Figure (2). The mass is also connected to a spring of spring constant (k=10 N/m). The other end of the spring is connected to a pivot. The unstretched length of the spring is 1 m. Find the equation of motion of the L=2 m system. m-5 kg 3 m k-10 N/m Figure 2 ww
- Derive the equations of motion using Lagrange's equations. (I, =mr? ) THE MOTION IS ON THE HORIZONTAL PLANE (NO GRAVITY) Trailer, mass M Cylinder, k1 mass m k219.In the situation shown in the figure, the spring constant is 2000 N/m and the length of the spring when it is neither stretched nor compressed is 80 cm. What is the length of the spring if this system rotates at 10 revolutions per second and if it is in a spaceship far away from all stars and planets (so there is no gravity)? 50 cm 1 kg 1 1 kg (There is no friction.)Q2) Consider a mass (m=5 kg) connected to a massless rod of length (L=2 m) to swing about point O, as shown in Figure (2). The mass is also connected to a spring of spring constant (k=10 N/m). The other end of the spring is connected to a pivot. The unstretched length of the spring is 1 m. Find the equation of motion of the L-2 m system. m-5 kg 3 m k=10 N/m