P13 006
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The document describes a scaffold with two cables holding it up next to a window. The scaffold has a mass of 60 kg and is 5 m long, suspended by cables 1.5 m and 3.5 m from the window. Taking torques about the far end of the scaffold, the tension F1 in the closer cable is calculated to be 8.4 x 102 N. Equilibrium of forces gives F1 + F2 = total weight, which is used with the known F1 to find F2 = 5.3 x 102 N.
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1) The document provides step-by-step work to solve multiple physics problems related to Batfink and a barrel rolling off a cliff.
2) It calculates the forces, acceleration, velocities, and distance needed to determine where to place a barrel catching device to safely catch the barrel.
3) The conclusion is that the barrel catching device should be placed 16.14 meters from the base of the cliff to catch the barrel falling with an initial velocity of 4.95 m/s.
7.60 ppt batfink energy
1) Batfink falls through a trap door and drops 20 meters, gaining 9800 J of kinetic energy just before hitting the ground.
2) He lands on a spring with a force constant of 700 N/m that compresses 3 meters, storing the 9800 J of energy.
3) Using the spring force and compression, the spring constant is calculated to be 700 N/m.
Uslides2
This document summarizes key concepts in physics related to motion, forces, energy, and sound waves. It covers Newton's laws of motion, definitions of speed, velocity, and acceleration. It also discusses gravitational force, momentum, work, power, heat transfer, and the properties of sound waves including wavelength, frequency, and how the velocity of sound varies with temperature. Key figures discussed include Aristotle, Galileo, Newton, and their contributions to the study and understanding of motion and force.
individual part
This document contains calculations for the dead loads, live loads, and ultimate loads on various beams and slabs in a building. It first calculates the loads on beam C-D, finding the total ultimate load to be 35.14 kN/m. It then calculates loads on beam C/2-3, with ultimate loads of 37.514 kN/m and 21.7 kN/m. Similar load calculations are provided for beams A1/3-4.1 and A1-B/3.1 and 4, finding ultimate point loads on these beams. The document includes details on the slab thicknesses, beam sizes, densities, and live load assumptions used in the calculations.
Capítulo 11 (5th edition)rewrweerww
The document contains solutions to physics problems involving kinematics, dynamics, work, energy, momentum, rotation, and torque. Key points include:
- Expressions are derived for the kinetic energy of translation and rotation, and the total kinetic energy of objects like disks and rings.
- Problems calculate accelerations, velocities, times, distances, energies, moments of inertia, and torques for objects rolling down inclines and rotating on an axis.
- Vector dot and cross products are used to find angles between vectors and rotational quantities like torque.
Phy i assign&answers_2011
This document contains physics questions and answers related to mechanics. Some key points:
- Question 11 discusses calculating the percent uncertainty in the volume of a spherical beach ball.
- Question 19 expresses a sum with the correct number of significant figures.
- Question 51 calculates the volume of the Moon and determines how many Moons would be needed to equal Earth's volume.
- Other questions cover topics like calculating components of velocity, displacement and time calculations for motion problems, projectile motion, and tensions in hanging objects. The document provides worked examples of physics equations and calculations.
Gaya
1. This document discusses concepts of physics related to acceleration, including calculations of net force, acceleration, displacement, time, and tension. It provides 10 examples of word problems and their step-by-step solutions involving these concepts.
2. The examples calculate values such as net force, acceleration, displacement, time, tension and maximum static friction based on given values of mass, force, weight, velocity, displacement and time.
3. Formulas used include calculations for net force, acceleration, velocity, displacement, time, tension, and static friction.
Sol35
1) The document analyzes the forces on beads inside a rising hoop to determine the minimum ratio of the beads' total mass (m) to the hoop's mass (M) needed for the hoop to rise.
2) It derives an expression for the outward force each bead exerts on the hoop as a function of the angle the bead has fallen (θ).
3) Taking the maximum of the total upward force from both beads, it determines this occurs when θ gives cosθ = 1/3, yielding a maximum upward force of 2mg/3.
4) For the hoop to rise, this maximum upward force must exceed the hoop's weight Mg. Equating
Sol35
1) The document analyzes the forces on two beads sliding down a rising hoop. It derives an expression for the outward force the beads exert on the hoop as a function of the angle fallen.
2) This force is maximized when the angle fallen is cos^-1(1/3), giving a maximum upward force of 2mg/3.
3) For the hoop to rise, this maximum upward force must exceed the weight of the hoop. This occurs when the ratio of bead mass to hoop mass is greater than 3/2.
Anschp23
This document summarizes Coulomb's law and provides example problems for calculating electrostatic force between point charges. Some key points:
- Coulomb's law defines the electrostatic force of attraction or repulsion between two point charges.
- A series of example problems calculate the force between various charge configurations using Coulomb's law and given charge magnitudes, distances, etc.
- The concept of resultant electrostatic force on a third charge placed between two other charges is introduced.
- Problems calculate this resultant force in different arrangements of multiple charges.
Problems on bearing capacity of soil
This document provides 10 examples of problems related to bearing capacity of foundations. The examples calculate bearing capacity using Terzaghi's analysis for different soil and foundation conditions, including cohesionless and cohesive soils, square and strip footings, and considering the water table depth. One example compares results to field plate load tests. The solutions show calculations for determining soil shear strength parameters, factor of safety, and safe bearing capacity.
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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 051
The document summarizes calculating torques and forces on a gymnastics beam system. It states that for computing torques, the axis is chosen at support 2 and clockwise is positive. Equating the torques leads to an expression for the upward force F1 at support 1 of 1163 N. It then states that balancing vertical forces gives an expression relating F1, F2, the masses and gravity, from which F2 is calculated to be 1740 N.
P13 050
The document describes calculating the forces acting on a beam attached to a hinge by a cable and balanced force. It finds that the cable exerts 88N of tension to balance the 50N force on the beam and no net torque. It then calculates the x and y components of the hinge force as 30N rightward and 97N upward to balance all the forces acting on the beam.
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 047
1) The document describes calculating the torque on a ladder leaning against a wall and finding the horizontal force needed to keep it from sliding.
2) Key values given include the applied force F, the weight W, and moment arms used to calculate torque.
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.
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 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.
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P13 006
- 1. 6. Let 1 = 1.5 m and 2 = 5.0 − 1.5 = 3.5 m. We denote the tension in the cable closer to the window as F1 and that in the other cable as F2 . The force of gravity on the scaffold itself (of magnitude ms g) is at its midpoint, 3 = 2.5 m from either end. (a) Taking torques about the end of the plank farthest from the window washer, we find mw g 2+ ms g 3 (80 kg) 9.8 m/s2 (3.5 m) + (60 kg) 9.8 m/s2 (2.5 m) F1 = = = 8.4 × 102 N . 1+ 2 5.0 m (b) Equilibrium of forces leads to F1 + F2 = ms g + mw g = (60 kg + 80 kg) 9.8 m/s2 = 1.4 × 103 N which (using our result from part (a)) yields F2 = 5.3 × 102 N.