Whenever we have two-dimensional vector problems in which no two vectors are parallel, the easiest method of solution is to pick a convenient coordinate system and project the vectors onto its axes. In this case the best coordinate system has one axis horizontal and the other vertical. We call the horizontal the
-axis
and the vertical the
-axis.
Solution
First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to draw a new free-body diagram showing all of the horizontal and vertical components of each force acting on the system.
When the vectors are projected onto vertical and horizontal axes, their components along those axes must add to zero, since the tightrope walker is stationary. The small angle results in
being much greater than
.
Consider the horizontal components of the forces (denoted with a subscript
):
The net external horizontal force
, since the person is stationary. Thus,
Now, observe
[link] . You can use trigonometry to determine the magnitude of
and
. Notice that:
Equating
and
:
Thus,
as predicted. Now, considering the vertical components (denoted by a subscript
), we can solve for
. Again, since the person is stationary, Newton’s second law implies that net
. Thus, as illustrated in the free-body diagram in
[link] ,
Observing
[link] , we can use trigonometry to determine the relationship between
,
,
and
.
As we determined from the analysis in the horizontal direction,
:
Now, we can substitute the values for
and
, into the net force equation in the vertical direction:
and
so that
and the tension is
Discussion
Note that the vertical tension in the wire acts as a normal force that supports the weight of the tightrope walker.
The tension is almost six times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is only a small fraction of the tension in the wire. The large horizontal components are in opposite directions and cancel, and so most of the tension in the wire is not used to support the weight of the tightrope walker.
If we wish to
create a very large tension, all we have to do is exert a force perpendicular to a flexible connector, as illustrated in
[link] . As we saw in the last example, the weight of the tightrope walker acted as a force perpendicular to the rope. We saw that the tension in the roped related to the weight of the tightrope walker in the following way:
We can extend this expression to describe the tension
created when a perpendicular force (
) is exerted at the middle of a flexible connector:
Questions & Answers
where we get a research paper on Nano chemistry....?
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest.
Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.?
How this robot is carried to required site of body cell.?
what will be the carrier material and how can be detected that correct delivery of drug is done
Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Receive real-time job alerts and never miss a matching job again
Source:
OpenStax, College physics arranged for cpslo phys141. OpenStax CNX. Dec 23, 2014 Download for free at http://legacy.cnx.org/content/col11718/1.4
Google Play and the Google Play logo are trademarks of Google Inc.
Notification Switch
Would you like to follow the 'College physics arranged for cpslo phys141' conversation and receive update notifications?