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- Mass of both blocks = 14kg. The friction coefficient for this bar is µ= 0.4. The spring constant k= 60 N/m. The spring is compressed 0.1 m and then released. d a Find: www 1. The magnitude of block B's acceleration 2. The magnitude of tension of the rope A B a=?A W = 58 lb block is pushed with a force that is defined by the equation F = (50+ s²) lb. The coefficient of kinetic friction of the block and surface is μ = 0.2. The spring is initially unstretched/uncompressed. s = 1 ft. F = f(s) k Values for the figure are given in the following table. Note the figure may not be to scale. Variable Value lb ft 01 30 degrees Using equations of motion in rectangular coordinates and principle of work and energy, N (s) = P = 16 Vfinal = The block is initially at rest. At the instant, the block has moved a distance of a. Find an equation for the normal reaction between the block and surface as a function of s, N(s). b. Determine final velocity of the block, Vfinal. c. Determine the power developed by the force at the instant, P, in horsepower. m Round your final answers to 3 significant digits/figures. Round your coefficients in equations to 3 significant digits/figures. lb S hp k ft S1. The PASCO human arm model is configured such that the cord representing the bicep is perfectly vertical and the forearm is at 90° (in the figure to the right, the cord is not quite vertical). A mass of 100 g is attached to the hand. Draw a free-body diagram on the figure to the right showing all forces which act on the forearm. The force of the bicep F on the arm The force of the humerus FH on the arm The weight of the forearm W The mass in the hand Wm 100 g Be careful to draw the force vectors with tails beginning at the point where the force is actually applied to the forearm. 2. Consider the free body diagram below. Determine the perpendicular component F̟ of the force F exerted by the biceps brachii on the forearm. Use the fact that cos 0 = H/B to write this component directly in terms of the humerus length H and the biceps length B. H 3. If the forearm is in equilibrium, then there is no angular acceleration and therefore the sum of the torques applied to the forearm must be…
- 1. An Atwood machine (shown below) consists of a pulley with two weights W₁ = 100 N and W₂ = = 200 N attached at the end of the string. T is the tension in the string and a is the acceleration of the system. If the tension and the acceleration satisfy the set of equations: T-W₁ = = 10a W₂-T = 20a Solve for the tension 7 and the acceleration a. It's okay if you do not understand the physics right now. Solve as a systems of linear equation problem. Solve for T and a: T= a = b WI 2 W2 N m/s²NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part A slender rod AB, with a weight of W, is attached to blocks A and B that move freely in the guides shown in the figure The constant of the spring is k. and the spring is unstretched when 0-0 Determine the value of 0 when W-75 lb, / 30 in, and k 3 lb/in. The value of isQ4-Two blocks A and B of weight 350 N and 600 N. respectively are connected by a cord that passes over a frictionless pulley, as shown in the Figure 8. Determine the force P to be applied to block B to produce the impending motion of block B down the plane. Fig. (8) A P₁-0.5 50⁰ Pa 0.4 B 30°
- 2. Figure 2 shows a quick return mechanism. The ram at C has a mass of 50kg and is accelerating to the right at 10 m/s². In addition there is a force resisting the motion of the ram of 1500N. Determine the magnitude and direction of the forces: (a) acting between the ram and guide (b) at the pin joint B (c) (d) (e) Find the torque about O. at the pin joint Q. at the slider A. BL 45° 71° 18.5° Figure 2. Quick Return Mechanism OA = 0.075m AQ = 0.162m BQ = 0.1m BC = 0.3mA 0.23 kg block is pressed up against a spring on a frictionless incline of angle 0 = 20.0° to the horizontal. The spring is initially compressed by x = 0.30 m from its equilibrium position (first arrow shown). When released from rest, the block travels along the incline a total distance (d + x) = 1.30 m until coming to rest again (where d = 1.00 m). What is the spring constant k? {Consider gravitational and spring potential energies; conservation of energy 'system'} O 11.5 N/m O 22.3 N/m O 80.9 N/m O 13.7 N/m O 30.2 N/m 8 LE R T K command command optionQ4/ Two identical spheres of radius 8 in. and weighing 2 lb on the surface of the earth are placed in contact. Find the gravitational attraction between them. Q5/ A person weighs 30 lb on the moon, where g = 5.32 ft/s2. Determine (a) the mass of the person and (b) the weight of the person on earth. Q6/ What is the weight in newtons of an object that has a mass of (a) 9 kg, (b) 0.06 kg.
- The top of a ladder weighing 20 kg and 6m long rests against a vertical wall and the foot of the ladder on a horizontal floor. For the wall, if = 0.15, and for the floor, f = O.2 O. How far up the ladder can a man weighing 60 kg go before the ladder begins to slip? Find the value of NB in terms of Y. * 60 Kg B -NB 0.15 kg X 20 kg 60° 0.2 NA NA1. In lecture I focused on a horizontal mass-spring system so we could ignore gravity, but in practice it's much easier to build a vertical mass-spring system. In the figure below, we can see a suspended spring with spring constant k both before and after a mass m is hooked to it. Y Yo yo is the equilibrium height of the spring (without the mass) and in the figure y = 0 is set at the ground. a) Using the coordinate system in the figure, show that Newton's second law for the hanging mass-spring system leads to the following differential equation: d²y dt² k (y - Yo) - g m b) Now define a new coordinate, y', related to y by a constant shift: y = y' + C, where C = constant. Show that if you choose the right value of C, Newton's second law is identical in form to a mass-spring system without gravity. What is the period of the system? c) Effectively, gravity just shifts the equilibrium position of the system, but the system still undergoes simple harmonic motion. What is the net force on the…Problem: 1F the cable that is connected block A A with an acceleration to OF 0.6 mlr Detormine the passibte Force in Newton OF block B, given its mars as 20 kg. Given: a= 0.6ms MB= 20 kg Raqd: FB %3D %3D Datum B. Datum+ soln: 25p+ h+ SA =L