4 Lab main

Help!!! Please answer part b correctly like part A. Please!!!!

Learning Goal: To learn to apply the method of joints to a truss in a systematic way and thereby find the loading in each member of the truss. In analyzing or designing trusses, it is necessary to determine the force in each member of the truss. One way to do this is the method of joints. The method of joints is based on the fact that if the entire truss is in equilibrium, each joint in the truss must also be in equilibrium (i.e., the free-body diagram of each joint must be balanced). Consider the truss shown in the diagram. The applied forces are P₁ = 630 lb and P₂ = 410 lb and the distance is d = 8.50 ft. Figure 1) 1 of 1 A E 30° d B D P₁ 30° d C

Learning Goal: To describe the shape and behavior of cables that are subjected to concentrated and distributed loads. Part A Structures often use flexible cables to support members and to transmit loads between structural members. Because a cable's weight is often significantly smaller than the load it supports, a cable's weight is considered negligible and, therefore, not used in the analysis. In this tutorial, cables are assumed to be perfectly flexible and inextensible. Thus, once the load is applied the geometry of the cable remains fixed and the cable segment can be treated as a rigid body. Cables of negligible weight support the loading shown. (Figure 1) If W, = 85.0 N , W, = 510 N, YB = 1.40 m, yc = 2.80 m, yp = 0.700 m, and zc = 0.850 m, find zg. Express your answer numerically in meters to three significant figures. > View Available Hint(s) VO AEoI vec IB = 2.048 m Submit Previous Answers X Incorrect; Try Again; 4 attempts remaining Part B Complete previous part(s) W2 O…

Having found FcD in the analysis of joint C, there are now only two unknown forces acting on joint D. Determine the two unknown forces on joint D. Express your answers in newtons to three significant figures. Enter negative value in the case of compression and positive value in the case of tension. Enter your answers separated by a comma.

Help!!! Please solve the value

Learning Goal: To calculate minor head losses and pressure drops for pipe fittings. Minor losses in pipe flow are the result of disruptions to the steady laminar or turblent flow in a pipe by entrances, bends, transitions, valves or other fittings. In general, calculating these losses analytically is too complex. However, all of these losses can be modeled using terms of the form h = KL where Kr, is called the loss coefficient and is determined experimentally. This coefficient relates the minor head loss to the velocity head for the flow. For expansions and contractions, the loss coefficient is calculated using nine the velocity for the smaller diameter pipe. The table below gives some representative values for various minor head losses. These are representative only, and a more complete table would account for different kinds of fittings and connections (like threaded or soldered). Fitting Well-rounded entrance ≥0.15 Flush entrance Re-entrant pipe Discharge pipe Sudden contraction (d₂…

Learning Goal: To use equilibrium to calculate the plane state of stress in a rotated coordinate system. In general, the three-dimensional state of stress at a point requires six stress components to be fully described: three normal stresses and three shear stresses (Figure 1). When the external loadings are coplanar, however, the resulting internal stresses can be treated as plane stress and described using a simpler, two-dimensional analysis with just two normal stresses and one shear stress (Figure 2). The third normal stress and two other shear stresses are assumed to be zero. The normal and shear stresses for a state of stress depend on the orientation of the axes. If the stresses are given in one coordinate system (Figure 3), the equivalent stresses in a rotated coordinate system (Figure 4) can be calculated using the equations of static equilibrium. Both sets of stresses describe the same state of stress. The stresses σx′and τx′y′ can be found by considering the free-body…

Learning Goal: To calculate the normal and shear stresses at a point on the cross section of a column. A column with a wide-flange section has a flange width b = 250 mm , height k = 250 mm , web thickness to = 9 mm , and flange thickness t; = 14 mm (Figure 1). Calculate the stresses at a point 65 mm above the neutral axis if the section supports a tensile normal force N = 2 kN at the centroid, shear force V = 5.8 kN , and bending moment M = 3 kN - m as shown (Figure 2). The state of stress at a point is a description of the normal and shear stresses at that point. The normal stresses are generally due to both internal normal force and internal bending moment. The net result can be obtained using the principle of superposition as long as the deflections remain small and the response is elastic. Part A- Normal stress Calculate the normal stress at the point due to the internal normal force on the section. Express your answer with appropriate units to three significant figures. • View…

Analysis and Interpretation of vertical Ground Reaction Forces. In this study, a volunteer was asked to walk on two force platforms under two different conditions. Condition 1- Normal Walk (NW), condition 2 - walking over an obstacle (ObsW). The right limb was the one to step on platform 1 in both conditions. The force platforms were used to measure the vertical ground reaction forces on the right limb. Data was collected on Vicon (Nexus) software and the attached graphs ( mean NW/ObsW, Mean (+-)1SD NW/ObsW) were created. Analyse ,Interpret the graphs and make a conclusion of the result attached.

Problem 1 Learning Goal: To be able to find the center of gravity, the center of mass, and the centroid of a composite body. A centroid is an object's geometric center. For an object of uniform composition, its centroid is also its center of mass. Often the centroid of a complex composite body is found by, first, cutting the body into regular shaped segments, and then by calculating the weighted average of the segments' centroids. Figure ←d→ x Part A IVE ΑΣΦ | 4 T, 1.610,0.5075 Submit An object is made from a uniform piece of sheet metal. The object has dimensions of a = 1.20 ft ,b= 3.74 ft, and c = 2.45 ft. A hole with diameter d = 0.600 ft is centered at (1.00, 0.600). Find z, y, the coordinates of the body's centroid. (Figure 1) Express your answers numerically in feet to three significant figures separated by a comma. ▸ View Available Hint(s) Previous Answers Provide Feedback vec • * Incorrect; Try Again; 4 attempts remaining ? 1 of 5 ft Review > Next > Activate Windows Go to…

Learning Goal: A column with a wide-flange section has a flange width b = 200 mm , height h = 200 mm, web thickness tw = 8 mm , and flange thickness tf = 12 mm (Figure 1). Calculate the stresses at a point 75 mm above the neutral axis if the section supports a tensile To calculate the normal and shear stresses at a point on the cross section of a column. normal force N = 2.9 kN at the centroid, shear force V = 4.6 kN, and bending moment M = 4.8 kN • m as shown (Figure 2). The state of stress at a point is a description of the normal and shear stresses at that point. The normal stresses are generally due to both internal normal force and internal bending moment. The net result can be obtained using the principle of superposition as long as the deflections remain small and the response is elastic. Part A - Normal stress Calculate the normal stress at the point due to the internal normal force on the section. Express your answer with appropriate units to three significant figures. > View…

Learning Goal: To use the principle of linear impulse and momentum to relate a force on an object to the resulting velocity of the object at different times. The equation of motion for a particle of mass m can be written as ∑F=ma=mdvdt By rearranging the terms and integrating, this equation becomes the principle of linear impulse and momentum: ∑∫t2t1Fdt=m∫v2v1dv=mv2−mv1 For problem-solving purposes, this principle is often rewritten as mv1+∑∫t2t1Fdt=mv2 The integral ∫Fdt is called the linear impulse, I, and the vector mv is called the particle's linear momentum.   A tennis racket hits a tennis ball with a force of F=at−bt2, where a = 1300 N/ms , b = 300 N/ms2 , and t is the time (in milliseconds). The ball is in contact with the racket for 2.90 ms . If the tennis ball has a mass of 59.7 g , what is the resulting velocity of the ball, v, after the ball is hit by the racket?

Help!!! Please answer all Correctly!!! Please

Learning Goal: To develop the ability to break a frame or machine down into subsystems and to determine the forces developed at internal pin connections. Frames and machines are systems of pin-connected, multiforce members. Frames are designed to support loads, whereas machines are designed to transmit or alter the effects of loads.For or machine to be in equilibrium, each member of the frame or machine system must be in equilibrium. Free-body diagrams of the overall system, as well as individual members, groups of members, and subsystems, must be drawn. Figure * B b 30° 26 E a- 30° K < H b 4 of 4 Submit Previous Answers Correct Note that the internal reactions at B are not included in the free-body diagram of the subsystem ABC. ▼ Part D A tractor shovel - The tractor shovel shown (Figure 4) carries a 540 kg load that has its center of mass at H. The shovel's dimensions are: a = 55.0 mm, b = 220 mm, c = 330 mm, d = 110 mm, and e = 385 mm. Find the reaction force at E. Assume that the…

After creating a decision matrix for two types of materials used to design a safety belt, an engineer assigns a weight of 4 to nylon for thickness and a weight of 5 to polyester for thickness. The engineer also assigns a weight of 4 to nylon for strength and a weight of 3 to polyester for strength. Polyester is more expensive than nylon. Describe which material would be preferable to use for the safety belt, if cost is prioritized as a criterion

Please help me solve this question for part A and part B  label final answer with correct units for P = and D =

Part C

Please do not copy other's work and do not use ChatGPT or Gpt4,i will be very very very appreciate!!! Thanks a lot!!!!!

Analysis and Interpretation of vertical Ground Reaction Forces. In this study, a volunteer was asked to walk on two force platforms under two different conditions. Condition 1- Normal Walk (NW), condition 2 - walking over an obstacle (OW). The right limb was the one to step on platform 1 in both conditions. The force platforms were used to measure the vertical ground reaction forces on the right limb. Data was collected on Vicon (Nexus) software and the attached graphs ( mean NW/OW, Mean (+-)1SD NW/OW) were created. Analyse and Interpret the graphs attached.

Statics Problem !!! Help me Part A , Part B , Part C!!!! Answer it this Problem Correctly!! Please give correct Solution

You are an engineer in a company that manufactures and designs several mechanical devices, and your manager asked you to help your customers. In this time, you have two customers, one of them wants to ask about internal combustion engines while the other requires a heat exchanger with particular specifications. Follow the parts in the following tasks to do your job and support your customers.Task 1:Your first customer asked for an internal combustion engine to use it in a designed car. Your role is to describe the operation sequence of different types of available engines, explain their mechanical efficiency, and deliver a detailed technical report which includes the following steps:STEP 1Describe with the aid of diagrams the operational sequence of four stroke spark ignition and four stroke compression ignition engines.STEP 2Explain and compare the mechanical efficiency of two and four-stroke engines.STEP 3Review the efficiency of ideal heat engines operating on the Otto and Diesel…

PART D

Learning Goal: To understand the derivation of the law relating height and pressure in a container.   In this problem, you will derive the law relating pressure to height in a container by analyzing a particular system.   A container of uniform cross-sectional area is filled with liquid of uniform density . Consider a thin horizontal layer of liquid (thickness ) at a height as measured from the bottom of the container. Let the pressure exerted upward on the bottom of the layer be and the pressure exerted downward on the top be . Assume throughout the problem that the system is in equilibrium (the container has not been recently shaken or moved, etc.). What is Fup , the magnitude of the force exerted upward on the bottom of the liquid?

Help!!! Please answer all Correctly!!! Please

Learning Goal: To be able to solve three-dimensional equilibrium problems using the equations of equilibrium. Part A As with two-dimensional problems, a free-body diagram is the first step in solving three-dimensional equilibrium problems. For the free-body diagram, it is important to identify the appropriate reaction forces and couple moments that act in three dimensions. At a support, a force arises when translation of the attached member is restricted and a couple moment arises when rotation is prevented.For a rigid body to be in equilibrium when subjected to a force system, both the resultant force and the resultant couple moment acting on the body must be zero. These two conditions are expressed as The J-shaped member shown in the figure(Figure 1) is supported by a cable DE and a single journal bearing with a square shaft at A. Determine the reaction forces Ay and Az at support A required to keep the system in equilibrium. The cylinder has a weight WB = 6.00 lb , and F = 1.40 lb…

Learning Goal: To use the principle of linear impulse and momentum to relate a force on an object to the resulting velocity of the object at different times. The equation of motion for a particle of mass m can be written as EF dv = ma = m dt By rearranging the terms and integrating, this equation becomes the principle of linear impulse and momentum: t2 ES F dt = m f dv = mv2 – mv1 V1 For problem-solving purposes, this principle is often rewritten as mvị +£ F dt = mv2 The integral F dt is called the linear impulse, I, and the vector my is called the particle's linear momentum. A tennis racket hits a tennis ball with a force of F = at – bt?, where a = 1200 N/ms , b = 500 N/ms? , and t is the time (in milliseconds). The ball is in contact with the racket for 3.05 ms . If the tennis ball has a mass of 62.6 g, what is the resulting velocity of the ball, v, after the ball is hit by the racket? Express your answer numerically in meters per second to three significant figures. • View Available…

The free body diagram must be drawn , its mandatory. Don't use chatgpt

Help!!! Answer all correctly with accurately. On the Part A. FBD make sure you Td , Tc, Ta values. Solve only Part A.

Learning Goal: To use transformation equations to calculate the plane state of stress in a rotated coordinate system. The normal and shear stresses for a state of stress depend on the orientation of the axes. If the stresses are given in one coordinate system (Figure 1), the equivalent stresses in a rotated coordinate system (Figure 2) can be calculated using a set of transformation equations. Both sets of stresses describe the same state of stress. In order to use the transformation equations, a sign convention must be chosen for the normal stresses, shear stresses, and the rotation angle. For the equations below, a positive normal stress acts outward on a face. A positive Try acts in the positive y-direction on the face whose outward normal is in the positive x-direction. The positive direction for the rotation is also shown in the second figure. The stresses in the rotated coordinate system are given by the following equations: στ σy + cos 20+Try sin 20 2 2 σετ συ = σy' cos 20-Try…