95.4.1 Eqns of Equil

Given: A 3500 lb motor and | 3500 lb plate, as shown, are in equilibrium and supported by three cables and d= 4 ft. 10 ft Find: Magnitude of the tension in each of the cables. В 4 ft `3 ft 3 ft 4 ft Plan: 1) Draw a free body diagram of Point A. Let the unknown force magnitudes be FB, Fc, F D . 2) Represent each force in the Cartesian vector form. 3) Apply equilibrium equations to solve for the 3 unknowns.

I did not understand this meaning

LTE I. 4:05 docs.google.com EE A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. The location of the rivet hole on line FG and line GH is..... /120 N 240 mm 200 N 70° 15 E D 50 mm 42 Nm 180 N 520 mm 60 N 40 Nm 180 mm F |B 50 mm 50 mm - 640 mm 0.259 mm, 0.298 mm O 0.298 mm, 0.259 mm O 260 mm, 300 mm 300 mm, 260 mm O

Please help explain the annotation?

3.8 Please help me with this problem with step by step clear explanation and needed correct solution, please.. I NEED ONLY THE CORRECT ANSWER Posting for a second time remember needed the correct answer ...

Q5 900 mm 700 mm An experimental device imparts a force of magnitude F=250 N to the front edge of the rim at A to simulate the effect of a slam dunk. B > The moment of the force F about point O is.... > The moment of the force F about point B is.... > The location of point C (from the base at O to the location on the ground where the force imparts zero moment) 300 mm 2 is... 3050 mm (Hint: The distance d from point O to point C is positive if C is to the right of O, and negative if to the left). 106.25 N.m CW, 141.7 N.m CW, 200 mm(to the right of O) 114.5 N.m CW, 157.2 N.m CW, 688 mm (to the left of O) 140 N.m CW, 190 N.m CW, 584 mm (to the right of 0) 140 N.m CW, 190 N.m CW, 584 mm (to the left of O)

An eye bolt is used to attach 3 cables to a steel plate. The tension in the three cables create F,=200 Ibf, F2=250 lbf, and F3=100 lbf with 0 = 30 degrees and p=27.1 degrees. If the eye bolt is in equilibrium, what is the x-component of the sum of other forces (reaction force) on the bolt? If you add up the three force vectors, the reaction force you are looking for will just be in the opposite direction to put the eye bolt in equilibrium. The x-direction is positive and to the right. Eye bolt rsteel plate Nut & Washer

Q2/ In the nucleus pulposus of an intervertebral disc, the compressive load is 1.5 times the externally applied load. In the annulus fibrosus, the compressive force is 0.5 times the external load. What are the compressive loads on the nucleus pulposus and annulus fibrosus of the L5-S1 intervertebral disc of a 90 kg man holding a 50 kg weight bar across his shoulders, given that 63% of body weight is distributed below the disc?

P235   - The gate below is an isosceles triangle with 3m side. What is the proper configuration, the triangulate upright or inverted, so that the forces are as minimal as possible?   - What would the forces be like in a circular 3m radius floodgate?

The following figure represents a bicep curl maneuver where the center of mass (CoM) of a 23 kg dumbbell is 0.29 m from the elbow joint (pivot). The mass of the arm is 5 kg and acts 0.17 m from the elbow joint, and the biceps tendon acts at a distance of 0.01 m from the elbow joint. A)The triceps produces a negligible force and the arm is at an angle of 15 degrees above the horizon. The given distances from the joint are the distances along the line that is 15 degrees above the horizon as shown in the figure. What magnitude of force in newtons (N) must the biceps muscle produce to hold the 23 kg mass in place (static equilibrium) b)

4.) A 20 kg block is suspended by 2 very thin ropes that meet at a 40° angle. What is the tension on the two ropes? AM ThN < Q口 Formula EQA D Untitled Notebook (2) - Untitled Notebook (2)- Satple Praloom s HI UNSM Ntbok2 Sanple Frirm 3 LnlT Sheet O O & C OBE 63 Atuoods Machine Block Suspended by a Strings 92=-a, (m. - mi). 9% = (mz +m.) Untitled Notebook (2) Tz = sine Block Sliding on a Romp Т. mg w/out friction: 9= gsine w/ friction:

clockwise moment = anticlockwise moment Centre of mass = x X N 0 100 N x 2 m essentials %3D Although weight is distributed through an object, it acts as a single, downward force from a So: the counterbalance should be placed 0.5 m from O. b Let F be the maximum load (in N). With this load on the crane, the counterbalance must produce its maximum moment about O. So it must be the greatest possible distance from 0- in other words, 1 m from it. As the crane is in equilibrium, the principle of moments applies: Taking moments about O: clockwise moment = point called the centre of mass (or centre of gravity). For an object to be stable when resting on the ground, its centre of mass must be over its base. If an object is pushed, and its centre of mass passes beyond anticlockwise moment 400 N X 1 m Fx 2 m %3D the edge of its base, it will F = 200 N %3D topple over. ment l balan So: the maximum load is 400 N. os pe ९SNO 1 In the diagram on the right, a plank weighing 120 N is supported by two…

Don't Use Chat GPT Will Upvote And Give Handwritten Solution Please

Help!

account_circle   Science PhysicsQ&A LibraryA child’s toy consists of a m = 31 g monkey suspended from a spring of negligible mass and spring constant k. When the toy monkey is first hung on the spring and the system reaches equilibrium, the spring has stretched a distance of x = 17.6 cm, as shown in the diagram. This toy is so adorable you pull the monkey down an additional d = 7.6 cm from equilibrium and release it from rest, and smile with delight as it bounces playfully up and down.  1. Calculate the speed of the monkey, ve, in meters per second, as it passes through equilibrium.  2. Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion, Etop, in terms of m, x, d, k, the maximum height above the bottom of the motion, hmax, and the variables available in the palette.   3. Calculate the maximum displacement, h, in centimeters, above the equilibrium position, that the monkey reaches. A child’s toy consists of a…

Question 1 The main forces acting on a flying aero plane are the thrust T, the weight W, and the aerodynamics forces namely the lift L and the drag D as shown in the figure Q1. The path of flight is shown by the x-axis. The angle between thrust and path is known as angle of attack and the angle between the horizontal and the path is known as flight path angle. The mass of the aero plane is 72000 kg. At an instant of equilibrium, the lift is 680 kN, the drag is 125 kN and the flight path angle is 6°. Determine the angle of attack and the thrust required to maintain a steady flight. Infer the effect of thrust on the aero plane's weight.

Given: Find: Fc ● FA A r60° FC=.79 lbs theta C w/ horizontal= 36° r=.98 in r B Written work: Draw the direction of F smallest force at A to balance moment: Fa = ##.## lbs X