Two wheels A and B with weights w and 7w, respectively, are connected by a uniform rod with weight. as shown below. The wheels are free to roll on the sloped surfaces. Determine the angle (in degrees) that the rod forms with the horizontal when the system is in equilibrium. Hint: There are five forces acting on the rod, which is two weights of the wheels, two normal reaction forces at points where the wheels make contacts with the wedge, and the weight of the rod. (Enter the smallest angle.) 15.175 B X° 30° 60° A
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- A massless cable is used to connect the center of mass of the cylinder A to the box B hanging over the pulley C. Take the coefficient of static friction as µs = 0.25 and that of kinetic friction as µk = 0.2. Assume that there is no slipping at the contact point D and solve the problem using Newton's second law to find: a) angular accelerations of the cylinder A and pulley C (aA and ac), b) tension forces in the cable on both sides of the pulley C (cylinder side is TA and box side is TB), and c) normal and rolling forces at D (Np and FD). d) Is the assumption of no slipping at the contact point D correct? Why or why not? Notes: 1) We are asuming that tension forces on two sides of the pulley C are not the same, but acceleration of the cable from A to B stays the same. 2) For a cylinder or disk of mass m and radius r, moment of inertia about its center is given as I = mr2/2. 3)Do not forget to sketch the FBD’s for the cylinder A, box B and pulley C. m, = 12 kg mc = 4 kg rc =0.2 m TA=0.5…A block with mass m = 5.00 kg slides down a surface inclined 36.9° to the horizontal (Figure 1). The coefficient of kinetic friction is 0.27. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 11.1 kg and moment of inertia 0.500 kg - m? with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.300 m from that axis. Part A What is the acceleration of the block down the plane? Express your answer in meters per second squared. ? a = 1.45 m/s? Figure 1 of 1 Part B What is the tension in the string? Express your answer in newtons. 5.00 kg ? T = 11.6 N 36.9 國You push on a rectangular door at the location of the knob (see diagram for top view). The door s mass is 62.6 kg, and its side-to-side width is 2.28 m. The knob is located 0.14 m from the right-hand side of the door. If you push with a force of 180 N, what will the be the door s angular acceleration as it swings on its hinges? (Note: the moment of inertia of the door of mass M and width x, swinging on its hinges, is (1/3) M x^2.) 5.68 rad/s^2 0.85 rad/s^2 3.55 rad/s^2 1.42 rad/s^2
- In part b) the moment which is in red color why doesn't equal -T( AP + Radius ) since they are perpendicular. Why did he use the components of TThe drawing below shows two forces acting on a rigid beam. The blue dot is the center of mass of the beam. The forces (red arrows) are equal in magnitude. Consider the net force on the beam and net torque about its center. Which of the following statements are true? Check ALL that apply. The net force is zero. The net force is not zero. The net torque is zero. O The net torque is not zero.You push on a rectangular door at the location of the knob (see diagram for top view). The door s mass is 71.9 kg, and its side-to-side width is 2.19 m. The knob is located 0.19 m from the right-hand side of the door. If you push with a force of 198 N, what will the be the door s angular acceleration as it swings on its hinges? (Note: the moment of inertia of the door of mass M and width x, swinging on its hinges, is (1/3) M x^2.) 3.45 rad/s^2 2.42 rad/s^2 5.18 rad/s^2 1.94 rad/s^2
- Hand written and clean answerYou push on a rectangular door at the location of the knob (see diagram for top view). The door s mass is 78.8 kg, and its side-to-side width is 2.47 m. The knob is located 0.19 m from the right-hand side of the door. If you push with a force of 198 N, what will the be the door s angular acceleration as it swings on its hinges? (Note: the moment of inertia of the door of mass M and width x, swinging on its hinges, is (1/3) M x^2.) 3.95 rad/s^2 0.79 rad/s^2 2.82 rad/s^2 1.13 rad/s^2Please help me by showing step by step.
- You push on a rectangular door at the location of the knob (see diagram for top view). The door s mass is 79.0 kg, and its side-to-side width is 1.91 m. The knob is located 0.11 m from the right-hand side of the door. If you push with a force of 101 N, what will the be the door s angular acceleration as it swings on its hinges? (Note: the moment of inertia of the door of mass M and width x, swinging on its hinges, is (1/3) M x^2.) 1.89 rad/s^2 1.13 rad/s^2 2.65 rad/s^2 0.79 rad/s^2You push on a rectangular door at the location of the knob (see diagram for top view). The door s mass is 73.5 kg, and its side-to-side width is 2.33 m. The knob is located 0.12 m from the right-hand side of the door. If you push with a force of 158 N, what will the be the door s angular acceleration as it swings on its hinges? (Note: the moment of inertia of the door of mass M and width x, swinging on its hinges, is (1/3) M x^2.) 2.63 rad/s^2 1.05 rad/s^2 3.68 rad/s^2 0.74 rad/s^2A uniform solid sphere of mass 6.0 kg and radius 10 cm is rolling without slipping along a horizontal surface with a speed of 4.9 m/s (this is the speed of the center of mass) when it starts up a ramp that makes an angle of 30° with the horizontal. 1st а) b) While the ball is moving up the ramp, find What is the maximum distance it can go up the ramp, measured along the surface of the ramp? 2nd? the acceleration (magnitude and direction) of its center of mass and i) ii) | 3rd 3] the friction force (magnitude and direction) acting on it due to the surface of the ramp. Moment of inertia of a solid sphere is I em =MR². ст