
Find a couple of M.

Answer to Problem 16.142P
The couple of M=-7.19N.m.
Explanation of Solution
Given information:
The attached rod A mass is=0.8kg
And length is=160mm
Rod BP mass is=1kg
Length is=200mm
Explanation:
First consider the unit
So, we have to find the vertical distance from point P and point A(y)
Here, we take the coordinate points as (0,0.1m)
Then the position of the vector
Coordinate points of the point P with the point B is
Then the position vector is
Then the coordinate points with the point P and point E as (0,-0.16m)
Position vector is
Here, the unit vector k as angular direction in clockwise is negative and it is positive when counter clockwise.
Angular velocity of the rod BP in vector form is
Velocity of the rod BP is
Formulate the velocity equation of point P
Compare the i terms from equation (1) And (2)
Compare the equation (1) and (2) for j terms. We get,
So, the angular acceleration is zero.
Formulate the acceleration of the rod BP
Find the acceleration of the point P to the point E
Formulate the acceleration at the point P
Substitute the above values
Based on equations (3) and (4), we get,
Find the weight of the rod AE is get by
Then the moment of inertial of the rod AE is
Similarly, find the weight of the rod BP is
Then the moment of inertial of the rod BP
Here, the mass of the centers of rod AE and BP are to be considered as G and H.
Then the reference for point G with point A as (0,-0.08m).
Here the position vector is −(0.08m)j.
Similarly, the point H with point B as the reference as (-0.0866m,-0.05m). Position vector is −(0.0866m)i-(0.05m)j
Acceleration of the point G is
Find the acceleration at point H
Based on above figure, consider the moment about point B.
The magnitude of the couple applied on the rod is BP. It is considered as -7.19N.m and acts in a clockwise direction.
Find the force exerted on rod AE.

Answer to Problem 16.142P
The force exerted by rod AE is F=52.757N.
Explanation of Solution
Considering the above diagram of free body of the rod AE
The force exerted by rod AE by block P is 52.75N. It is acted upon the left direction.
Want to see more full solutions like this?
Chapter 16 Solutions
VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS
- Solve this problem and show all of the workarrow_forwardThe 4-lbs piece of putty is dropped 12 ft onto the 16-lbs block initially at rest on the two springs, each with a stiffness k = 5 lbs/in. Calculate the additional deflection d of the springs due to the impact of the putty, which adheres to the block upon contact.arrow_forwardSolve this probem. Draw the diagram and show how the moments are calculated in all of the directionsarrow_forward
- Solve this problem and show all of the workarrow_forwardThe 4-lbs piece of putty is dropped 12 ft onto the 16-lbs block initially at rest on the two springs, each with a stiffness k = 5 lbs/in. Calculate the additional deflection d of the springs due to the impact of the putty, which adheres to the block upon contact.arrow_forwardA converging elbow (see the figure below) turns water through an angle of 135° in a vertical plane. The flow cross section diameter is 400 mm at the elbow inlet, section (1), and 200 mm at the elbow outlet, section (2). The elbow flow passage volume is 0.2 m³ between sections (1) and (2). The water volume flowrate is 0.1 m³/s and the elbow inlet and outlet pressures are 140 kPa and 90 kPa. The elbow mass is 11 kg. Calculate the (a) horizontal (x direction) and (b) vertical (z direction) anchoring forces required to hold the elbow in place. D₂- Section (1)- D₁ = 400 mm 135° 200 mm Section (2) (a) Fx= i 20809.96 N (b) Fz= i -120265 Narrow_forward
- 2: A rectangular aluminum block is loaded uniformly in three directions. The loadings are as follows:a 50 kN total resulting compressive load in the x-direction, a 200 kPa uniformly distributed tensile load in they-direction, and a 0.03 MN total resulting tensile load in the z-direction. The block has the following dimensions:L = 1 m, b = 20 cm, h = 350 mm. Use E = 70 GPa and ν = 0.25.Determine the strain in the x and y axes respectively. For the strain in the y-direction to be equal to 0, how much uniformly distributed load inthe surface of y-direction should be added? (+ for tensile, - for compressive) Answers: 5 -1.122 x10-5 / 3 decimals 6 5.102 x10-6 / 3 decimals 7 -0.357 MPa / 3 decimalsarrow_forwardA spherical balloon with a diameter of 9 m is filled with water vapour at 200˚C and 200 kPa. Determine the mass of water in the balloon. The R value for water is 0.4615 kJ/kg∙K.arrow_forwardCorrect answers are written below. Detailed and correct solution only with fbd. I will upvote. 1: A 3 m alloy shaft fixed at one end has a torsional shearing stress capacity of 55 MPa. Due to improper fabrication, its cross-sectionalarea has become irregularly shaped. Its effective polar moment of inertia has become 2 x10-7 m4, and the maximum torque stress acts at 7.5 cm fromthe center of the shaft.[1]: If the shaft is to be replaced by a properly manufactured solid circular shaft that has a maximumshearing stress capacity of 70 MN/m2, what is the minimum diameter required so it can withstand the sameload? [2]: Calculate the thickness of a hollow circular shaft with the same outside diameter calculated initem [1] that can carry the same load. Limit the maximum shearing stress of the hollow circular shaft to0.09 GPa.Determine the angle of twist on the free end of the shaft. Use G = 150 x103 GPa. [3]: Use the solidcircular shaft from [1] and use the hollow circular shaft from [2].…arrow_forward
- two closed 1 m3 chambers are filled with fluid at 25˚C and 1 atm. One is filled with pure carbon dioxide and one is filled with pure water. Only considering the weight of the fluids, which chamber is heavier?arrow_forwardCorrect answers are written below. Detailed and correct solution only with fbd. I will upvote. 1: A 3 m alloy shaft fixed at one end has a torsional shearing stress capacity of 55 MPa. Due to improper fabrication, its cross-sectionalarea has become irregularly shaped. Its effective polar moment of inertia has become 2 x10-7 m4, and the maximum torque stress acts at 7.5 cm fromthe center of the shaft.[1]: If the shaft is to be replaced by a properly manufactured solid circular shaft that has a maximumshearing stress capacity of 70 MN/m2, what is the minimum diameter required so it can withstand the sameload? [2]: Calculate the thickness of a hollow circular shaft with the same outside diameter calculated initem [1] that can carry the same load. Limit the maximum shearing stress of the hollow circular shaft to0.09 GPa.Determine the angle of twist on the free end of the shaft. Use G = 150 x103 GPa. [3]: Use the solidcircular shaft from [1] and use the hollow circular shaft from [2].…arrow_forwardPlease can you assist me with the attached question. Many thanks.arrow_forward
- Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University PressMechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSONThermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
- Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEYMechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage LearningEngineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY





