Problem P4.15 Gripper C of the industrial robot is accidentally subje.pdf

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Problem P4.15 Gripper C of the industrial robot is accidentally subjected to a 60 -lb side load directed perpendicular to BC (Figure P4.15). The lengths of the robot\'s links are AB = 22 in. and BC = 18 in. By using the moment components method, determine the moment of this force about the center of joint A. Solution It is given that AB is making an angle of 15 with horizonatal and BC is making an angle of 40 with AB therefore the angle of BC with the horizontal will be 55 and 60 lb force is perpendicular to the BC therfore it will make an angle of 35 with the horizontal.So we will breakinto components along horizontal and vertical as shown in the figure. Now we will calculate the perpendicualr distance between the forces and point A. Perpendicular distance of point A from 60Sin35 will be equal to = AD +BE = 31.57 Perpendicular distance of point A from 60Cos35 will be equal to = BD + CE = 20.43 Now from the right angle triangle ABD and BCE we will calculate the value of AD = 21.25 , BE = 10.32, BD = 5.69 , CE = 14.74 Now the moment at point A due to 60Sin35 will be = 60sin35*31.57 (anticlockwise) Due to 60Cos35 = 60Cos35*20.43 (Anticlockwise) Since the nature of moment of both the forces is same therefore they will be added Net moment at point A = 2090.58 lb-in.

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points) r False (f statement is false, correct it that it is true) a) Polymers make up an exciting class of biomaterials Polyethylene is a wonderful example of a simple ceramic that can be used in multiple applications by modifying its chemical structure. It tends to be bioactive when used by itself as an implant, but it becomes even more bioactive when 10 vol% Calcium Carbonate is added to it. Teflon can be used as a vascular graft material due to its low coefficient of friction. This is achieved because C atoms are replaced by F atoms in its structure. Collagen is the premiere polymer that is produced in our bodies. There are about 15 types of collagen and this polysaccharide is very important in bone matrix formation and tissues and organs. b) A vacancy is a point defect that occurs when an atom is missing from a lattice position. A perfect metal crystal would exhibit ductile behavior. Amorphous crystals have long range order which can be measured by x-ray diffraction. X-ray crystal structure determination and the iridescent colors in nacre, butterfly wings, and beetle exoskeletons are caused by similar phenomena that involves the transmission of visible light. Metals tend to be harder than polymers, ceramics tend to be stronger than polymers, and metals tend to be harder than ceramics. Diamond is a unique material that exhibits metallic bonding. One of its more unusual features is that although it has a very high electrical conductivity, it is a poor conductor of heat. c) H bonding is a very important bond type in biological materials such as water and DNA. H bonding is responsible for the anomalous freezing behavior of water in which its Solution a) true b) false because diamond is good conductor of heat because of the strong covalent bonding and low phonon scattering. Thermal conductivity of natural diamond was measured to be about 22 W/(cm·K), which is five times more than copper. Monocrystalline synthetic diamond enriched to 99.9% the isotope 12C has the highest thermal conductivity of any known solid at room temperature: 33.2 W/(cm·K).diamond is a good electrical insulator, having a resistivity of 100 G·m to 1 E·m c) false density decrease when transform solid to liquid thays why ice use to float at water surface d) TRUE rapid cooling producing small crystals, slower cooling producing larger ones e) FALSE (100) plane physically intersects the Y and Z axes and its perpendicular to both plane (010)and(020) thanks.

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Please read the description below and identify which region the text is referring to by dragging and dropping the name of the region into the bins provided. Historically, this region has suffered from extreme poverty, which continues to this day. Growth has been effectively zero since the 1960?s. Problems such as political instability, poor public health and a lack of effective institutions have all contributed to its stagnation. Until the middle of the last century, these countries were relatively poor. Beginning in the mid 1970?s, real GDP per capita growth has averaged 6% per year. The growth was achieved, in part, because of high levels of investment spending in the development of human and physical capital, and rapid technological progress. In the early 1990s, this region was in the middle of substantial social and economic reform. Countries in this region have experienced variable growth rates, depending on their ability to adapt to the modern market economy. In the early 20th century, this region was reasonably prosperous. Since that time, however, growth has stagnated, owing to government instability, banking failures and run-away inflation. Recently, some countries in the region have begun to grow more consistently, with one country becoming a powerhouse of world economic development. Solution Statement 1 represents the situation of Sub-Africa. The region has suffered from poverty for a long time now. Economic growth has been absent ever since they were free from colonial rulers. Statement 2 describes the situation of Asian tigers - Hong Kong, Singapore, South Korea and Taiwan. Much of growth in these countries was enabled by export led industrialization. Statement 3 describes situation of European Transition Economies where end of Socialism and integration with rest of the world took place. The growth record haven\'t been really impressive since 1990s/ Statement 4: South America suits the best here. Brazil is the nation that is becoming the powerhouse of economic development..

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Problem 7: U2 has a concert that starts in 17 minutes and they must all cross a bridge to get there. All 4 men begin on the same side of the bridge. You must help them across to the other side. It is night time. There is one flashlight. A maximum of 2 people can cross at the same time. Any party who crosses, either 1 or 2 people, must have the flashlight with them. The flashlight must be walked back and forth; it cannot be thrown, etc. Each band member walks at a different speed. A pair must walk together at the i-ate of the slower man?s pace: * Bono: 1 minute to cross * Edge: 2 minutes to cross * Adam: 5 minutes to cross * Larry: 10 minutes to cross For example: if Bono and Larry walk across first, 10 minutes have lapsed when they get to the other side of the bridge. If Larry then returns with the flashlight, a total of 20 minutes have passed and you have already failed the mission. You have to get everybody in only 17 minutes! Remember that there is no trick behind this. Can you help them? No, you do not have to use group theory. Solution The trick is here to think and act. Bono and Edge crosses the bridge in two minute. They go to nearby place and gets two more flash lights in 2 minutes. Bono returns with two flash lights in 1 minute and asks Adam and Larry to take one flash light first. They take 10 minutes to cross. Total time elapsed = 2+2+1+10 = 15 Now Bono starts and crosses in one minute Thus within 17 minutes they crossed the bridge and ready for the concert..

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Problem 6 Like most civil engineering materials, asphalt concrete has upsides and downsides in terms of sustainability. The Calkins text outlines many of these in Chapter 8. Often, asphalt\'s downsides can be mitigated by alternative practices or materials Provide a discussion of one potential environmental issue related to asphalt (A)Discuss one aspect of traditional asphalt concrete that is unsustainable (environmentally, economically, etc.) (B) Discuss two potential means/methods for mitigating or eliminating the issue described in (A) Solution Asphalt concrete is extensively used to build roads. Asphalt concrete mainly contains asphalt which is used as a binder and some other aggregate of minerals. Asphalt layer is laid upon a base of gravel layer to construct a stiff and dependable road. Asphalt can be categorized into different categories depending on the laying temperature of the asphalt. advantages of asphalt :- Asphalt is used for highways having high volume of traffic. These roads are lower in cost and produce lower noise making it more suitable for regular usage with safety. It also offers higher traffic speed and is easy to repair as well disadvantages of asphalt :- However, Asphalt roads are less durable and become soft in hot weathers. Higher temperatures can soft down the asphalt binder making the roads softer. Thus a heavy load on such soft surface can deform the surface. However, very cold temperature can cause the contraction of asphalt resulting in the formation of cracks. Good maintenance can increase the durability of the roads. It also reduces the cost of repairing A and B he networks that tie our world together are indeed vast constructions, none more so than the roadways that tie our cities, towns and villages together. Roadways the world over are predominantly made of asphalt, an impervious, petroleum-based substance that includes aggregate such as sand or gravel. This same product is used in the installation of the expansive parking lots that characterize suburban development.What makes asphalt all the more problematic is the fact that water cannot pass through it. As a result, asphalt – in addition to paving the world over – not only dries out the ground below it, but gradually warms the environment in which it is used. Beyond ecological and aesthetic complexities, asphalt’s chemical makeup also represents a very real threat in both the short and long term. For one thing, asphalt has very high thermal mass properties through which it can retain heat generated from the sun. Said heat is slowly released from pavement over time, creating the heat island effect, which is responsible for increasing temperatures in some cities by as much as ten per cent relative to surrounding rural areas. Because asphalt is manufactured out of oil, moreover, it releases substantial volatile organic compounds (VOCs) during its production, including carcinogens benzene and toluene. Due to the intricate molecular structure of such compounds.

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Problem 24 A dernetme sig with a trander function Problem 2.4 A discrete-time signal with a transfer function H1(2) is connected in a feedback configuration as shown in figure 7.7(c). The feedback loop system H2(z) consists of a simple one-sample delay. Compute the overall transfer function H(z) Solution Overall transfer function = H1(z) / 1+H1(z)H2(z) H2(z) = z-1 so H(z) = {z/(4z2-4z+2)} / [1+ 1/(4z2-4z+2)] = z / 4z2-4z+2 +1 so overall H(z) = z / 4z2-4z+3.

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Problem 14-1A On January 1, 2014, Geffrey Corporation had the following stockholders\' equity accounts Common Stock ($20 par value, 60,000 shares issued and outstanding) Paid-in Capital in Excess of Par-Common Stock Retained Earnings $1,200,000 200,000 600,000 During the year, the following transactions occurred. Declared a $1 cash dividend per share to stockholders of record on February 15, payable March 1 Paid the dividend declared in February Announced a 2-for-1 stock split. Prior to the split, the market price per share was $36 Declared a 10% stock dividend to stockholders of record on July 15, distributable July 31, On July 1, the market price of the stock was $13 per share Issued the shares for the stock dividend Declared a $0.50 per share dividend to stockholders of record on December 15, payable January 5, 2015 Determined that net income for the year was $350,000 Feb. 1 Mar. 1 Apr. 1 July 1 31 Dec. 1 31 Journalize the transactions and the closing entries for net income and dividends. (Credit account titles are automatically indented when amount is entered. Do not indent manually. If no entry is required, select \"No Entry\" for the account titles and enter 0 for the amounts. If no entry is required, select \"No entry\" for the account titles and enter 0 for the amounts.) Date Aocount Titles and Explanation Debit Credit Feb. 1 Mar. 1 Apr. 1 July 1 uly 31 Dec. 1 Dec. 31 (To close net income) (To close stock dividends) (To close cash dividends) Solution Journal Entries: Date Accounts / Explanations Debit Credit Feb. 1 Cash Dividend $ 60,000.00 Dividend Payable $ 60,000.00 (Being cash dividend declared = 60000 Shares * $1) Mar. 1 Dividend Payable $ 60,000.00 Cash $ 60,000.00 (Being cash dividend cash ) Apr. 1 No entry (New shares = 60000*2 = 120000 Shares @ $20/ 2 =$10 Par value) July. 1 Stock Dividend $ 156,000.00 Dividend Distributable $ 156,000.00 (Being Stock Dividend Declared = 120000 *10% = 12000 Shares*$13 ) July. 31 Dividend Distributable $ 156,000.00 Paid in capital in excess of par- Common Stock (156000-120000) $ 36,000.00 Common Stock (12000 Shares * $10) $ 120,000.00 (Being shares issues for dividend ) Note : total number of shares after Stock Div = 120000+12000 = 132000 Dec. 1 Cash Dividend $ 66,000.00 Dividend Payable $ 66,000.00 (Being cash dividend declared = 132000 Shares * $0.50) Dec. 31 Income Summary $ 350,000.00 Retained earnings $ 350,000.00 (Being net income closed) Retained earnings $ 126,000.00 Cash Dividend (60000+66000) $ 126,000.00 (Being cash dividend closed) Retained earnings $ 156,000.00 Stock Dividend $ 156,000.00 (Being cash dividend closed) Journal Entries: Date Accounts / Explanations Debit Credit Feb. 1 Cash Dividend $ 60,000.00 Dividend Payable $ 60,000.00 (Being cash dividend declared = 60000 Shares * $1) Mar. 1 Dividend Payable $ 60,000.00 Cash $ 60,000.00 (Being cash dividend cash ) Apr. 1 No entry (New shares = 60000*2 = 120000 Shares @ $20/ 2 =$10 Par value) July. 1 Stock Dividend $ 156,000.00 Div.

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Problem 11-13 Ratio analysis-comprehensive problenm Presented here are summarized data from the balance sheets and income statement WIPER, INC. Condensed Balance Sheets December 31, 2014, 2013, 2012 in milions) 2014 2013 2012 Current assets Other assets 677 $ 891 736 2,413 1,920 1,719 3,090 $2,811 $2,455 803 $ 710 827 918 $ 562 Current liabilities Long-term liabilities Owners equity 982 1,521 1,0071,026 $3,090$2,811$2,455 3.090 $2.811 $2,455 WIPER, INC Selected Income Statement and Other Data For the year Ended December 31, 2014 and 2013 (in millions) 2014 . : 2013 Income statement data Sales Operating income (EBIT) Interest expense Net income $3,050$2,913 310 65 187 296 84 192 Other data Average number of common shares outstanding Total dividends paid 41.3 46.7 s 50.0S52.3 Required a. Calculate returm on investment. based on net income e and average totaf assets, for 2014 and 2 Solution Return on Investment = Net Income x 100 / Average Assets. Average Assets = Assets at beginning + Assets at end For 2014: 192 x 100 / 2,950.5 = 6.5% For 2015: 187 x 100 / 2,633 = 7.1% Days’ Sales in Receivables = Average receivables / Sales / 365 309 / 3060 / 365 = 309 / 8.38 = 37 days. Debt Ratio = Total debt / Capital Employed 2014 = 2083 / 2528 = 82.4% 2013 = 1785 / 2008 = 88.9% Debt Equity Ratio = Long-term debt / Owners’ equity 2014 = 1521 / 1007 = 151% 2013 =982 / 1026 = 96% Times Interest Earned Ratio = EBIT / Total Interest Payable 2014 =296 / 84 = 3.5 times 2013 =310 / 65 = 4.8 times..

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