'

Search results

Found 547 matches
NTU method - the heat transferred between the fluids (q)

The Number of Transfer Units (NTU) Method is used to calculate the rate of heat transfer in heat exchangers (especially counter ... more

Ideal rocket equation (Tsiolkovsky rocket equation)

The Tsiolkovsky rocket equation, or ideal rocket equation describes the motion of vehicles that follow the basic principle of a rocket: a ... more

Kepler's Third Law

In astronomy, Kepler’s laws of planetary motion are three scientific laws describing the motion of planets around the Sun.

1.The orbit of a ... more

Distance between two points (three-space)

Distance is a numerical description of how far apart objects are. In analytic geometry, the distance between two points of the xyz-plane in three-space, ... more

Central processing unit power consumption

Central processing unit power dissipation or CPU power dissipation is the process in which central processing units (CPUs) ... more

Gearing reduction ratio

Harmonic Drive is the brand name of strain wave gear trademarked by the Harmonic Drive company, and invented in 1957 by C.W. Musser.

It is very ... more

Height of a trapezoid in relation with the sides

Trapezoid is a convex quadrilateral with only one pair of parallel sides. The parallel sides are called the bases of the trapezoid and the other two sides ... 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/

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/

Flux (as a single scalar)

Flux is two separate simple and ubiquitous concepts throughout physics and applied mathematics. Within a discipline, the term is generally used ... more

...can't find what you're looking for?

Create a new formula