P13 025
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The document summarizes calculations to determine forces in a cable-strut system. It defines key angles between components as 15 and 45 degrees. It uses the torque equation about the hinge to find the torque T as 6.63x103 N by canceling the unknown boom length. Horizontal force Fx is then calculated as 5.74x103 N using T and the 30 degree cable angle. Vertical force Fy is calculated as 5.96x103 N using torque T, weights and gravity.
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3) The torque equation is set to zero and rearranged to solve for the horizontal ground force Fgx for different values of F and W.
Learningobject1
1) The document provides information about a simple pendulum experiment with a basketball hanging from a string. It asks the learner to calculate various properties of the pendulum's motion.
2) The learner is asked to find the net force, tension, restoring force, acceleration, angular frequency, and displacement of the pendulum at different points in its swing.
3) Conceptual questions ask how increasing the string length or mass would affect the angular frequency of the pendulum's oscillation.
Bef 2
This document discusses beams supported on an elastic foundation. It begins by introducing the Winkler foundation model and defining short, medium, and long beams based on the parameter βL. It then provides solutions for the deflection, slope, bending moment and shear force of an infinite beam under a point load. The document also discusses beams supported by discrete elastic supports and beams subjected to a distributed load segment. It provides examples calculating deflection, bending stress, and pressure for specific beam problems.
Shear Force and Bending Moment
Shear force and bending moment diagrams are constructed for a beam subjected to various loads. The shear force is maximum at supports and zero at points of inflection, while the bending moment is maximum at points of inflection and zero at supports. For the given beam under a uniform distributed load and two concentrated loads, the shear force and bending moment diagrams are drawn showing the variations along the length of the beam. Key points of zero shear force and maximum bending moment are identified.
Capitulo 9, 7ma edición
1) A point charge moving through an electric field is shown to follow a parabolic trajectory given by z = -1.5×1010t2 m. At t = 3 μs, its position is found to be P(0.90, 0, -0.135) m.
2) A point charge moving through a uniform magnetic field follows a circular path. The equations of motion are derived and solved, giving the position, velocity, and kinetic energy at t = 3 μs.
3) Forces on current loops and filaments in various magnetic field configurations are calculated.
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P13 062
The document summarizes the forces acting on a symmetrical structure with diagonal bars connecting top and bottom horizontal bars. It determines that the diagonal bars each exert a force of T√2 at a 45 degree angle on the horizontal bars, placing them in tension with a force T. As the structure is symmetrical under 90 degree rotations, all horizontal and diagonal bars must be in tension with a force T.
P13 059
A balance has arms of length 1 and 2, and masses m1 and m2. The equations for torque equilibrium when the balance is balanced are m1*1=m2*2 and m1=m2*2. Dividing the equations and solving for the unknown mass gives the result that m = √(m1*m2).
P13 058
The document describes calculating the forces on an object hanging from a hinge. It finds the catching force to be 42 N by setting up a torque equilibrium ratio using the object's weight, distance from the hinge, and distance below the hinge. It then uses vertical force equilibrium to find the hinge force is 66 N by subtracting the catching force from the object's weight.
P13 057
The document analyzes the forces at a knot holding a 5 kg mass at a 30 degree angle. It represents the static friction force and weight of the 5 kg mass as the sides of a right triangle. It finds that the static friction force equals the weight of the 5 kg mass multiplied by the tangent of 30 degrees. For the static friction force to be at its maximum, it must equal the coefficient of static friction multiplied by the weight of the 10 kg object. Equating these expressions allows one to calculate that the coefficient of static friction equals 0.29.
P13 056
The document shows a calculation of the torque equilibrium for a rod with forces applied at different distances from the fulcrum. The perpendicular force F⊥ required for equilibrium is calculated to be 8.7 N based on a 40 N force applied 2.6 cm from the fulcrum and the distance to the fulcrum being 12 cm.
P13 053
1) The document describes calculating the critical condition for a cube to begin sliding when a force P is applied to its top surface.
2) It sets static friction equal to its maximum value and models the normal force as acting at the bottom corner below where P is applied.
3) Equating the torques due to static friction and the force of gravity yields the critical condition that the coefficient of friction μ must equal L/2h.
P13 052
The cube has side length l and volume V=l^3. The document calculates the pressure p required using the formula p=BΔV/V, where B is a constant and ΔV is the change in volume. It shows that ΔV/V ≈ 3Δl/l. Plugging in the values for a cube where the side length increases from 85.0 cm to 85.5 cm, it calculates the required pressure p is approximately 2.4 × 10^9 N/m^2.
P13 048
This document describes calculating torques and forces on a beam hinged at one end. It finds that for a 12m beam with a 400N tension and angles of 40 and 50 degrees, the weight on the beam is 671N. Equilibrium equations show the horizontal hinge force is 400N of tension and the vertical force is approximately 670N of weight.
P13 046
This document describes the tensions in two strings (T and T') that are supporting a weight (W). It provides the force equilibrium equations relating the horizontal and vertical forces of T, T', and W. Solving these simultaneous equations yields that T = 15 N and T' = 29 N.
P13 045
The force exerted on the beam is 7900 N. To find the required cross-sectional area, the formula F/A = Su/6 is used, giving an area of 9.5 x 10-4 m2. Taking the square root of this area gives the thickness of 0.031 m.
P13 044
A steel cable supporting a 670 kg load experiences an elastic stretch proportional to its free length. For a cable length of 12m, the calculated stretch is 0.0008m. For a longer cable length of 350m, the calculated stretch increases to 0.023m due to the cable's increased free length.
P13 043
The document calculates the strain, stress, and Young's modulus of a rope used for rock climbing. It determines that the strain is 1.9 × 10−3 based on the rope's stretched length. The stress is calculated to be 1.3 × 107 N/m2 by relating the force of a 95 kg climber to the rope's cross-sectional area. Finally, it divides the stress by the strain to determine the rope's Young's modulus is 6.9 × 109 N/m2.
P13 042
The document summarizes the calculation of:
1) The volume of sand contained within a cylinder and cone of given dimensions is 6.78 cubic meters.
2) The weight of the sand is calculated to be 1.20 × 105 Newtons based on the volume and density.
3) An equation for stress as a function of radius is derived, with stress proportional to 13r + 40000 pascals.
4) The force on the cylinder is integrated and calculated to be 1.04 × 105 Newtons.
5) The fractional reduction in calculated force compared to measured weight is 13%.
P13 041
The document describes the equations used to determine the angle and tension required to move a log just before it begins to slide. It provides three equations setting the horizontal and vertical forces equal to the maximum static friction and torque, which are solved simultaneously to find the angle is 51 degrees when the coefficient of static friction is 0.8. The tension is then calculated to be 0.64 times the weight of the log.
P13 040
The flat roof of a building has an area of 870 square meters. The volume of material directly above an underground tunnel 60 meters deep is 52,200 cubic meters. With the material having a density of 2,800 kg/m3, the total mass supported by steel columns is 1.46 x 108 kg. The weight supported is 1.4 x 109 Newtons, requiring around 75 columns each able to support 400,000 Newtons over 960 square meters.
P13 039
1) The forces exerted by wires A (FA) and B (FB) on a log are related by the stretching of the wires based on their original lengths (LA and LB), cross-sectional areas, and Young's modulus.
2) Equating the total force on the log (FA + FB - mg) to 0 allows solving for FA and FB in terms of the known parameters.
3) FA is calculated to be 866 N and FB is then calculated from the total force equation to be 143 N.
P13 038
The brick compresses the two cylinders equally by an amount Δ. This means the force exerted by each cylinder is proportional to its cross-sectional area and elastic modulus. The ratio of the forces is 4:1. Combining this with the equation that the total force equals the weight gives the individual forces as 4W/5 and W/5. Computing torques about the center of mass confirms the 4:1 ratio of the distances of the forces from the center.
P13 037
The shear stress on an aluminum rod due to the weight of a 1200 kg object hanging from it is calculated to be 6.5 × 106 N/m2 based on the force applied, cross-sectional area of the rod, and radius. The shear modulus of the aluminum is used to calculate the vertical deflection of the rod under this stress as 1.1 × 10−5 m by relating the shear stress, length of the rod, and modulus.
P13 036
The Young's modulus (E) of a material was calculated from its stress-strain curve to be 7.5 × 1010 N/m2 based on a stress of 150 × 106 N/m2 and strain of 0.002. The linear range of the curve extended to approximately 2.9 × 108 N/m2, which represents the yield strength of the material where plastic deformation begins.
P13 034
(1) The document discusses the balancing of blocks on a table to determine the maximum height possible given constraints on the center of mass of each block.
(2) For part (a), the maximum height is found to be 9L/8 when the center of mass of each upper block must be above the edge of the lower block.
(3) For part (b), a more subtle condition is considered where forces are exerted between blocks; here the maximum height is found to be 7L/6.
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P13 025
- 1. 25. (a) We note that the angle θ between the cable and the strut is 45◦ − 30◦ = 15◦ . The angle φ between the strut and any vertical force (like the weights in the problem) is 90◦ − 45◦ = 45◦ . Denoting M = 225 kg and m = 45.0 kg, and as the length of the boom, we compute torques about the hinge and find M g sin φ + mg 2 sin φ T = . sin θ The unknown length cancels out and we obtain T = 6.63 × 103 N. (b) Since the cable is at 30◦ from horizontal, then horizontal equilibrium of forces requires that the horizontal hinge force be Fx = T cos 30◦ = 5.74 × 103 N . (c) And vertical equilibrium of forces gives the vertical hinge force component: Fy = M g + mg + T sin 30◦ = 5.96 × 103 N .