Concept explainers
Figure (a) shows the cross section of a column that uses a structural shape known as
Trending nowThis is a popular solution!
Chapter 9 Solutions
International Edition---engineering Mechanics: Statics, 4th Edition
- A nonuniform horizontal bar of mass mmm is supported by two massless wires against gravity. The left wire makes an angle a with the horizontal, and the right wire makes an angle B. The bar has length L. (Figure 1) What is the position of the center of mass of the bar, measured as distance x from the bar's left end? Express your answer in terms of a, ß, and L. a ·L .mm. -X Barrow_forwardThe following diagram corresponds to questions 1 to 3. A beam rests on two sharp edges as shown on the image. The beam has a length of 7,50 m and a mass of 4,25 kg. Object 1 has a mass of 1,50 kg; object 2 has a mass of 2,35 kg. Point P is 0,50 m form the center of the beam. d 2 Pl CG 1. How much is the torque done by object 1 around point P? A. 55,2 N*m B. 7,36 N*m С. 62,5 N*m D. Object 1 doesn't exert a torque.arrow_forwardConsider the U-beam section shown in the figure below. The beam section splits into 3 segments (assuming it to be made of uniform, material and the mass of the vertical segment is double the mass of the horizontal segments). Find the y-coordinate of the center of mass from the bottom of the beam section (in cm). 120 cm 30 cm 180 cm + 40 cm 30 cm Origin 120 cm O a. 120.0 Ob. 177.5 Oc 214.3 O d. 52.5 O e None of the choicesarrow_forward
- 3) A point mass is attached to one end of a beam levered against a spike which acts as the pivot. The point mass weighs 250N and the beam has a uniform mass of 10kg and is 8m long. Parts a and b Part c Fapplied 3 of 6mov 1m 8m Krist 0 = 60° Scale 250 N Scale a. If a scale placed beneath the point mass reads 250N, what is the magnitude and direction for the force at the pivot point? Direction: North, 90° from x-axis Magnitude to k* 250N = 2500 N b. For the same scenario, how far must the pivot point be located relative to the left edge of the beam? (Hint, both the boxes weight and the force from the scale act at the farthest right edge of the beam.) www.acta 0672nos onl c. The pivot is then moved to be 6m from the left edge of the beam which now makes a 60 angle with the floor. How much downward force would be needed at the left-most edge of the beam to reduce the reading on the scale to zero? (Hint: the beam is still in static eq. when the scale reads zero.)arrow_forward4. For the three cylinders shown in the figure, determine: a. the reaction in the contact of cylinder A and the horizontal surface b. the reaction in the contact of cylinder C and the vertical wall (Final answers in N, 3 decimal places) ■ 100 43 N C 37 N 80 A 85 N 120 B 365 All the dimensions are in mmarrow_forwardFor the beam I-section below, calculate the second moment of area about its centroidal x-axis (Ixxcentroid), where b1 = 12.50 mm, b2= 2.50 mm, b3 = 25.00 mm, d1 = 4.25 mm, d2 = 19.50 mm and d3= 6.50 mm. Give your answer in mm4 to two decimal places. b1 d2 ➜ → b₂ b3 d₁arrow_forward
- A chemistry graduate student is designing a pressure vessel for an experiment. The vessel will contain gases at pressures up to 80.0 MPa. The student's design calls for an observation port on the side of the vessel (see diagram below). The bolts that hold the cover of this port onto the vessel can safely withstand a force of 4.70 MN, pressure vessel bolts aide port Calculate the maximum safe diameter w of the port. Round your answer to the nearest 0.1 cm.arrow_forwardThe following diagram corresponds to questions 1 to 3. A beam rests on two sharp edges as shown on the image. The beam has a length of 7,50 m and a mass of 4,25 kg. Object 1 has a mass of 1,50 kg; object 2 has a mass of 2,35 kg. Point P is 0,50 m form the center of the beam. CG 1. How much is the torque done by object 1 around point P? A. 55,2 N*m B. 7,36 N*m C. 62,5 N*m D. Object 1 doesn't exert a torque. 2. How much is the torque done by the force of gravity of the beam around point P? A. 20,8 N*m N-m ך177 .B C. 156 N*m D. 313 N*m 3. If you needed to cancel the nomal forces of the two objects, where you should place object 2? The axis of rotation is point P. A. 3,30 m from point B. 9,18 m from point P C. 5,69 m from point P D. 3,62 m from point Parrow_forwardProblem 4: Values: a = 300 mm b = 75 mm c = 100 mm Use the graphical method to break F into components parallel to bars AB and BC. If the component parallel to bar AB is 7 kN, what is the magnitude of F? Comments: B This is a perfect example of why the graphical method is useful - we are asked to resolve a force into components that are not perpendicular to one another. It's difficult to apply regular old sine and cosine (i.e. trig) for this because we're not dealing with right triangles. Instead we will use the law of sines and/or the law of cosines. • Start by drawing a vertical line to represent the vector F. Now draw a line parallel to AB passing through the top of F, and another line, also parallel to AB, through the bottom. Repeat for BC. You should have a parallelogram divided into two triangles by the resultant vector F. Pick one of the triangles to work with, use geometry to figure out the angles, and then apply the law of sines to compute the magnitude of F. Remember, these…arrow_forward
- S00 Ib In the fig. shown, compute the ff: (16-18) the resultant using cosine law (force polygon) 60 R = 35 (19-20) the angle of the R measured 500 lb cW from the x- axis.arrow_forwardQUESTION 2 Question 2 A cross-section of a beam is shown in Figure Q2. If the shear force in this section is V = 125 KN, determine the value and the location of the maximum shear stress in the section. In Figure Q2, a = 30 mm and the origin of the coordinate system is at centroid of the cross section. 7 y= Z= A a AY S= 20 4a mm; mm; O Figure Q2 Answer The vertical coordinate (y-coordinate; the y-axis serves as the axis of symmetry of the cross- section.) and horizontal coordinate (z-coordinate) of the location where the maximum shear stress occurs in the section are ← a The vertical distance from the location where the maximum shear stress occurs in the section to the bottom side (AB cross section can be calculated as Distance = mm (units: mm) 3a Second moment of area The second moment of area employed in the equation to calculate maximum shear stress can be calculated as I₂ = a (units: mm²) Shear stress The second moment of area employed in the equation to calculate maximum shear…arrow_forwardConsider the forces acting on the bracket shown in (Figure 1). Figure Submit VG ΑΣΦ 459 (F₁)z. (F1), (F₁) = Determine the coordinate angle y for F2.(Figure 1) Express your answer in degrees to three significant figures. Submit Request Answer (F2), (F2), (F2), = /30 ΑΣΦ | 11 60% F₂ = 600 N Submit Request Answer VAX 1 vec vec X Incorrect; Try Again; 4 attempts remaining F₁ = 450 N Previous Answers Request Answer Express the force F₁ as a Cartesian vector. Enter the components of the force separated by commas. Express your answers in newtons to three significant figures. ——| ΑΣΦ | |t | vec 195| | S 1 of 1 A @ GERIC ? Express the force 12 as a Cartesian vector. Enter the components of the force separated by commas. Express your answers in newtons to three significant figures, N ? ? Narrow_forward
- International Edition---engineering Mechanics: St...Mechanical EngineeringISBN:9781305501607Author:Andrew Pytel And Jaan KiusalaasPublisher:CENGAGE L