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Proper motion (right ascension)

Proper motion is the astronomical measure of the observed changes in the apparent places of stars or other celestial objects in the sky, as seen from the ... more

Beta Angle

The beta angle is a measurement that is used most notably in spaceflight. The beta angle determines the percentage of time an object such as a spacecraft ... more

Hubble's Law

Hubble’s law is the name for the observation in physical cosmology that: (1) objects observed in deep space (extragalactic space, ~10 megaparsecs or ... more

Declination of the Sun (simplified)

The position of the Sun in the sky is a function of both time and the geographic coordinates of the observer on the surface of the Earth. As the Earth ... more

Declination of the Sun

The position of the Sun in the sky is a function of both time and the geographic coordinates of the observer on the surface of the Earth. As the Earth ... more

Absolute Magnitude of a Star - with luminosity distance

Absolute magnitude is the measure of a celestial object’s intrinsic brightness. It is the hypothetical apparent magnitude of an object at a standard ... more

Absolute Magnitude of a Star - with distance modulus

Absolute magnitude is the measure of a celestial object’s intrinsic brightness. It is the hypothetical apparent magnitude of an object at a standard ... more

Absolute Magnitude of a Star - with parallax

Absolute magnitude is the measure of a celestial object’s intrinsic brightness. It is the hypothetical apparent magnitude of an object at a standard ... more

Screw Thread Engagement

Bolted joints are one of the most common elements in construction and machine design. They consist of fasteners that capture and join other parts, and are ... more

Tunnel Ionization - DC

Tunnel ionization is a process in which electrons in an atom (or a molecule) pass through the potential barrier and escape from the atom (or molecule). In ... more

Worksheet 289

Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust T , for the four-rocket propulsion system shown in the Figure below. The sled’s initial acceleration is 49 m/s 2, the mass of the system is 2100 kg, and the force of friction opposing the motion is known to be 650 N.

A sled experiences a rocket thrust that accelerates it to the right.Each rocket creates an identical thrust T . As in other situations where there is only horizontal acceleration, the vertical forces cancel. The ground exerts an upward force N on the system that is equal in magnitude and opposite in direction to its weight,w.The system here is the sled, its rockets, and rider, so none of the forces between these objects are considered. The arrow representing friction ( f ) is drawn larger than scale.
Assumptions: The mass of the Sled remains steady throughout the operation

Strategy

Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in the figure.

Solution

Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. Since we have defined the direction of the force and acceleration as acting “to the right,” we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with

Force (Newton's second law)

Fnet is the net force along the horizontal direction, m is the rocket’s mass and a the acceleration.

We can see from the Figure at the top, that the engine thrusts add, while friction opposes the thrust.

Subtraction

Tt is the total thrust from the 4 rockets, Fnet the net force along the horizontal direction and Ff the force of friction.

Finally, since there are 4 rockets, we calculate the thrust that each one provides:

Division

T is the individual Thrust of each engine, b is the number of rocket engines

Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/

Tensile Stress Area

Bolted joints are one of the most common elements in construction and machine design. They consist of fasteners that capture and join other parts, and are ... more

Radial acceleration in circular motion ( related to period)

Uniform circular motion, that is constant speed along a circular path, is an example of a body experiencing acceleration resulting in velocity of a ... more

Worksheet 306

Calculate the force the biceps muscle must exert to hold the forearm and its load as shown in the figure below, and compare this force with the weight of the forearm plus its load. You may take the data in the figure to be accurate to three significant figures.


(a) The figure shows the forearm of a person holding a book. The biceps exert a force FB to support the weight of the forearm and the book. The triceps are assumed to be relaxed. (b) Here, you can view an approximately equivalent mechanical system with the pivot at the elbow joint

Strategy

There are four forces acting on the forearm and its load (the system of interest). The magnitude of the force of the biceps is FB, that of the elbow joint is FE, that of the weights of the forearm is wa , and its load is wb. Two of these are unknown FB, so that the first condition for equilibrium cannot by itself yield FB . But if we use the second condition and choose the pivot to be at the elbow, then the torque due to FE is zero, and the only unknown becomes FB .

Solution

The torques created by the weights are clockwise relative to the pivot, while the torque created by the biceps is counterclockwise; thus, the second condition for equilibrium (net τ = 0) becomes

Force (Newton's second law)
Torque
Force (Newton's second law)
Torque

Note that sin θ = 1 for all forces, since θ = 90º for all forces. This equation can easily be solved for FB in terms of known quantities,yielding. Entering the known values gives

Mechanical equilibrium - 3=3 Torque example

which yields

Torque
Addition

Now, the combined weight of the arm and its load is known, so that the ratio of the force exerted by the biceps to the total weight is

Division

Discussion

This means that the biceps muscle is exerting a force 7.38 times the weight supported.

Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/

Ball Screw - Preload Drag Torque

A ball screw is a mechanical linear actuator that translates rotational motion to linear motion with little friction. A threaded shaft provides a helical ... more

Drag equation ( for fluids)

Drag (sometimes called air resistance, a type of friction, or fluid resistance, another type of friction or fluid friction) refers to forces acting ... more

Gompertz–Makeham Law of Mortality

The Gompertz–Makeham law states that the human death rate is the sum of an age-independent component (the Makeham term, named after William Makeham) and an ... more

Force (Newton's second law)

In physics, a force is any influence which tends to change the motion of an object.In other words, a force can cause an object with mass to change its ... more

Jounce

In physics, jounce or snap is the fourth derivative of the position vector with respect to time, with the first, second, and third derivatives being ... more

Gyrofrequency

If the magnetic field is uniform and all other forces are absent, then the Lorentz force will cause a particle to undergo a constant acceleration ... more

Radius of Inertial circle ( by Coriolis effect)

In physics, the Coriolis effect is a deflection of moving objects when they are viewed in a rotating reference frame.
An air or water mass moving with ... more

Worksheet 302

In the wheelbarrow of the following figure the load has a perpendicular lever arm of 7.50 cm, while the hands have a perpendicular lever arm of 1.02 m.(a) What upward force must you exert to support the wheelbarrow and its load if their combined mass is 45.0 kg? (b) What force does the wheelbarrow exert on the ground?


(a) In the case of the wheelbarrow, the output force or load is between the pivot and the input force. The pivot is the wheel’s axle. Here, the output force is greater than the input force. Thus, a wheelbarrow enables you to lift much heavier loads than you could with your body alone. (b) In the case of the shovel, the input force is between the pivot and the load, but the input lever arm is shorter than the output lever arm. The pivot is at the handle held by the right hand. Here, the output force (supporting the shovel’s load) is less than the input force (from the hand nearest the load), because the input is exerted closer to the pivot than is the output.

Strategy

Here, we use the concept of mechanical advantage.

Force (Newton's second law)
Mechanical Advantage - Law of Lever
Subtraction

Discussion
An even longer handle would reduce the force needed to lift the load. The MA here is:

Division

Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/

1st Equation of Motion - Linear Velocity

In mathematical physics, equations of motion are equations that describe the behaviour of a physical system in terms of its motion as a function of ... more

Annualisation of logarithmic retururn

In finance, return is a profit on an investment. It comprises any change in value, and interest or dividends or other such cash flows which the investor ... more

Tractive Force - Steam locomotives

As used in mechanical engineering, the term tractive force can either refer to the total traction a vehicle exerts on a surface, or the amount of the total ... more

Energy of damped harmonic motion

Damped harmonic motion is a real oscillation, in which an object is hanging on a spring. Because of the existence of internal friction and air resistance, ... more

Rod and piston-to-stroke ratio

In a reciprocating piston engine, the connecting rod or conrod connects the piston to the crank or crankshaft. Together with the crank, they form a simple ... more

Drift Velocity (with current and conductor section area)

The drift velocity is the average velocity that a particle, such as an electron, attains in a material due to an electric field. It can also be referred to ... more

Drift Velocity

The drift velocity is the average velocity that a particle, such as an electron, attains in a material due to an electric field. It can also be referred to ... more

Worksheet 296

(a) Calculate the buoyant force on 10,000 metric tons (1.00×10 7 kg) of solid steel completely submerged in water, and compare this with the steel’s weight.

(b) What is the maximum buoyant force that water could exert on this same steel if it were shaped into a boat that could displace 1.00×10 5 m 3 of water?

Strategy for (a)

To find the buoyant force, we must find the weight of water displaced. We can do this by using the densities of water and steel given in Table [insert table #] We note that, since the steel is completely submerged, its volume and the water’s volume are the same. Once we know the volume of water, we can find its mass and weight

First, we use the definition of density to find the steel’s volume, and then we substitute values for mass and density. This gives :

Density

Because the steel is completely submerged, this is also the volume of water displaced, Vw. We can now find the mass of water displaced from the relationship between its volume and density, both of which are known. This gives:

Density

By Archimedes’ principle, the weight of water displaced is m w g , so the buoyant force is:

Force (Newton's second law)

The steel’s weight is 9.80×10 7 N , which is much greater than the buoyant force, so the steel will remain submerged.

Strategy for (b)

Here we are given the maximum volume of water the steel boat can displace. The buoyant force is the weight of this volume of water.

The mass of water displaced is found from its relationship to density and volume, both of which are known. That is:

Density

The maximum buoyant force is the weight of this much water, or

Force (Newton's second law)

Discussion

The maximum buoyant force is ten times the weight of the steel, meaning the ship can carry a load nine times its own weight without sinking.

Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/

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