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Normal force for a sinking object settles on the solid floor

When a sinking in a fluid object settles on the solid floor, it experiences a normal force.

... more

Tsiolkovsky rocket equation - acceleration based

The Tsiolkovsky rocket equation, classical rocket equation, or ideal rocket equation is a mathematical equation that describes the motion of vehicles that ... 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/

Acceleration of a particle in an electric field

The electric field is a component of the electromagnetic field. It is a vector field, and it is generated by electric charges or time-varying magnetic ... 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/

Gravitational Acceleration

Gravity gives weight to physical objects and causes them to fall toward the ground when dropped.
If Μ is a point mass or the mass of a sphere with ... more

Thrust

Thrust is a reaction force described quantitatively by Newton’s second and third laws. When a system expels or accelerates mass in one direction, the ... more

Thrust (with cross section area)

Thrust is a reaction force described quantitatively by Newton’s second and third laws. When a system expels or accelerates mass in one direction, the ... more

Cyclotron resonance frequency

A cyclotron is a type of particle accelerator in which charged particles accelerate outwards from the center along a spiral path. The particles are held to ... more

Tension to restrain a floating object

Archimedes’ principle states that “Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the ... 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/

Newton's second law Newton's second law (constant-mass system)

The second law states that the net force on an object is equal to the rate of change of its linear momentum in an inertial reference frame. The second law ... 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/

Worksheet 316

Calculate the change in length of the upper leg bone (the femur) when a 70.0 kg man supports 62.0 kg of his mass on it, assuming the bone to be equivalent to a uniform rod that is 45.0 cm long and 2.00 cm in radius.

Strategy

The force is equal to the weight supported:

Force (Newton's second law)

and the cross-sectional area of the upper leg bone(femur) is:

Disk area

To find the change in length we use the Young’s modulus formula. The Young’s modulus reference value for a bone under compression is known to be 9×109 N/m2. Now,all quantities except ΔL are known. Thus:

Young's Modulus

Discussion

This small change in length seems reasonable, consistent with our experience that bones are rigid. In fact, even the rather large forces encountered during strenuous physical activity do not compress or bend bones by large amounts. Although bone is rigid compared with fat or muscle, several of the substances listed in Table 5.3(see reference below) have larger values of Young’s modulus Y . In other words, they are more rigid.

Reference:
This worksheet is a modified version of Example 5.4 page 188 found in :
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/

Angular Acceleration

Torque, moment, or moment of force is the tendency of a force to rotate an object about an axis, fulcrum, or pivot.
Moment of inertia is the mass ... more

Weight

In science and engineering, the weight of an object is usually taken to be the force on the object due to gravity.
In Newtonian physics the weight is ... more

Vertical Forces at the wheels for a bicycle (rear wheel)

Though longitudinally stable when stationary, a bike may become longitudinally unstable under sufficient acceleration or deceleration. The normal ... more

Vertical Forces at the wheels for a bicycle (front wheel)

Though longitudinally stable when stationary, a bike may become longitudinally unstable under sufficient acceleration or deceleration. The normal ... more

Vis-Viva Equation

In astrodynamics, the vis viva equation, also referred to as orbital energy conservation equation, is one of the fundamental equations that govern the ... more

Clairaut's theorem

Clairaut’s theorem is a general mathematical law applying to spheroids of revolution. The formula can be used to relate the gravity at any point on ... more

Specific Impulse

Specific impulse (usually abbreviated Isp) is a way to describe the efficiency of rocket and jet engines. It represents the force with respect to the ... 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

The power to thrust ratio - rocket propulsion

Spacecraft propulsion is any method used to accelerate spacecraft and artificial satellites. Space propulsion or in-space propulsion exclusively deals with ... more

Tsiolkovsky rocket equation as function of payload

The Tsiolkovsky rocket equation, classical rocket equation, or ideal rocket equation is a mathematical equation that describes the motion of vehicles that ... more

Miles Equation

In 1954, Miles developed his version of this equation for GRMS as he was researching fatigue failure of aircraft structural ... more

Gravity Acceleration by Altitude

The gravity of Earth, which is denoted by g, refers to the acceleration that the Earth imparts to objects on or near its surface due to gravity. In SI ... more

Specific Impulse by weight

Specific impulse (usually abbreviated Isp) is a measure of the efficiency of rocket and jet engines. By definition, it is the impulse delivered per unit of ... more

Specific Impulse by weight - with mass flow rate

Specific impulse (usually abbreviated Isp) is a measure of the efficiency of rocket and jet engines. By definition, it is the impulse delivered per unit of ... more

Vis-Viva Equation with standard gravitational parameter

In astrodynamics, the vis viva equation, also referred to as orbital energy conservation equation, is one of the fundamental equations that govern the ... more

Uniform Circular Motion position (Y - coordinate)

In physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with ... more

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