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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.Mason is is trying to keep a door open using a door stopper. That door is solid, and it has a mass of 18.3kg. The door closer is connected 26.2cm away from the hinge, and pulls with a force of 123N. The door stopper provides a static friction force of 13.2N, and a kinetic friction force off 11.8N. Find the angular acceleration (as a vector) of the door.2 Your answer is partially correct. Try again. In the figure, a slab of mass m, = 40 kg rests on a frictionless floor, and a block of mass m, = 9 kg rests on top of the slab. Between block and slab, the coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.40. A horizontal force F of magnitude 102 N begins to pull directly on the block, as shown. In unit-vector notation, what are the resulting accelerations of (a) the block and (b) the slab? u = 0- (a) Number T-7.4 -7.4 Units m/s^2 (b) Number i+ Units m/s^2 -0.88 Click if you would like to Show Work for this question: Open Show Work SHOW HINT LINK TO TEXT LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM VIDEO MINI-LECTURE 10:50 PM search ENG 4/4/2021 ond 111Pgup) pri sc DOUSE insert dele home break 13) 23 bac 3 5. 4 R.
- Show the complete solution. Put notes on steps. The wooden block (mass m = 0.6165 kg) is released from rest at A by a compressed spring (compressed length 0.6 m, undeformed length 1 m, spring constant k = 150 N/m). The block is allowed to slide through the rough horizontal surface (A to B), then along the smooth circular ramp (B to C, central angle 0 = 45°, until the block is released after point C. Calculate the speed of the block at points B and C. Also, what is the magnitude of the normal force exerted to the block just before the block leaves the ramp? Neglect the geometry of the block.NOTE: Use Work-Energy Method to solve for the speeds; use Force-Mass-Acceleration (FMA) Method to compute for the normal force.The two ramps pictured here each have a region of length L with kinetic friction coefficient uk and are otherwise frictionless. Identical masses m start at identical heights up the ramps, with the same initial velocity down the ramp. How do their final speeds compare when they reach the far right-hand side? (The answer requested here is qualitative and does not require detailed calculation). L m m Vi #0 Vi O Mass is moving faster in #1 The masses are moving at the same speed. Mass is moving faster in #2 O There is not enough information to decide. L Mk#0 1 2 m m Vf,1 V₁,210:57 68 Practice: A + In Fig 3-16, the two boxes have identical masses of 40 kg. Both experience a sliding friction force with Hk = 0.15. Find the acceleration of the boxes ng sin 30 30 and the tension in the tie cord. Fig. 3-16 Kimberly S •.. Niea Sama. & Princess Di.. & Kimberly S
- A small object A is held against the vertical side of the rotating cylinderical container of raidus r due to entrifugal action the coefficient of astatic friction between the object and the container is , determine the expression for the maximum rotational rate co of the container which will keep the object from slipping out the vertical side. Fanpas (Fry: Mos FMS.N Fn-man +N=mrw +N=mYw² {fy.g +4₁ (norw')= ndgT my F Two blocks connected by a rope of negligible mass are being dragged by a horizontal force F. Suppose that F=68.0 N, mị = 12.0kg, m2 = 18.0kg, and %3D the coefficient of kinetic friction between each block and the surface is 0.1. (a) Draw a free-body diagram for each block. (b) Find the acceleration of the system. (C) Determine the tension T of the rope between the blocks.Net torque, moment of inertia, and constant angular acceleration The radius of the roll of paper shown below is 7.6 cm and its moment of inertia is I= 3.3 x 10-3 kg-m². The roll is initially at rest, then a downward force F = 3.5 N is exerted on the end of the roll for 1.3 s, as shown. A constant frictional torque (Tf.) with magnitude 0.11 N-m that slows the clockwise rotation acts on the roll from the moment it starts rotating until the moment it stops rotating. After the force F is removed (at the end of 1.3 s), the roll slows and eventually comes to a stop due to tr. Assume that the paper is so thin that the radius of the roll remains constant while it is unrolling. (a) Calculate the total length of paper that unrolls from the moment the force F was applied until the moment that the roll stops moving. (answer: 7.3 m) (b) The thickness of the paper is about 0.01 cm. Is the assumption that the radius of the roll remained constant justified? That is, estimate the fractional decrease in…
- Harlgw Cylinder エ:Mg The 65 kg man at right is trapped inside a section of large pipe. If that's not bad enough, the pipe begins to roll, from rest, down a 35 m long, 180 incline! The pipe has a mass of 180 kg and a diameter of 1.2 m. (Assume the man's presence inside the pipe has a negligible effect on the pipe 's rotational inertia.) The coefficient of friction between the pipe and the ground is (0.5, 0.4). -RO Motion Information Free-Body Diagram Event 1: Event 2: 「り t = 2=35m 02 = V2 = 02 = SF a = a2 = (moM) 9 aj = a2 = Mathematical Analysis34 at g sino 4:4F - (mt M) gcoS 2: (m + M)g sino - 'F = (mr M) a zて-Ix two eluntions %3D こ :zC =I« : 'FR = MR*< = MR (4) F = Ma to Sdve for FFL IV-95 Chele tun maxThe slender, homogeneous rod shown in the figure below is 3 m in length and has a mass of 12 kg. At the instant shown, the rod has an angle to the horizontal of 60-degrees and an angular velocity of 2 rad/s clockwise. Both walls are smooth and provide negligible friction. At the instant shown, what are the reaction forces at A and B? What is the angular acceleration of the rod's center of mass?kihetic friction), f = µN. f is directed opposite to the relative velocity. Example 3.4 Block and Wedge with Friction A block of mass m rests on a fixed wedge of angle 6. The coefficient of friction is u. (For wooden blocks, u is typically in the range of 0.2 to 0.5.) Find the value of 6 at which the block starts to slide, and the acceleration X when it slides. In the absence of friction, the block would slide down the plane; hence the friction force f points up the plane. With the coordinates shown, we have m.X=W sin 6-f and SHOT ON MI A2 MI DUAL CAMERA mỹ = N-W cos 0 2021/4/2 19:27