Question

In the figure, a particle of mass m1 = 0.94 kg is a distance d = 32 cm from one end of a uniform rod with length L = 4.2 m and mass M = 5.9 kg. What is the magnitude of the gravitational force ->F on the particle from the rod?

Answer 1

@Mertsj @ash2326 @saifoo.khan I think they can help you.

Answer 2

Lol ok, thanks! @Mertsj Can you help?

Answer 3

I do not know physics. Sorry

Answer 4

Thats ok thanks anyways :| lol

Answer 5

ANy figure given?

Answer 6

yes! http://edugen.wileyplus.com/edugen/courses/crs4957/art/qb/qu/c13/fig13_41.gif

Answer 7

I figure that I have to relate the masses some how. and i know im supposed to integrate a force but idk where to start -_-

Answer 8

Is that a wrong link? Can't see any figure.

Answer 9

Can you take a screenshot?

Answer 10

I don't think the length of the rod matters. If it doesn't,
(.94 * 5.9 * 9.8)/(.032)^2
(mass1 * mass2*gravity)/distance^2

Answer 11

They need an edit button. .32, not .032

Answer 12

figure

Answer 13

Thats what I was thinking too but it turns out not to be a simple problem. @Kitsuneinabox

Answer 14

Since the rod is an extended object, we cannot
apply Equation 13-1 directly to find the force.
Instead, we consider a small differential element of
the rod, of mass dm of thickness dr at a distance
r from 1 m . The gravitational force between dm and 1 m is
1 1
2 2
dF Gm dm Gm (M / L)dr
r r
= = ,
where we have substituted dm = (M / L)dr since mass is uniformly distributed. The
direction of dF
..
is to the right (see figure). The total force can be found by integrating
over the entire length of the rod


this is what my solution manual says, bleahhh

Answer 15

I'll take a look in about an hour if I can.

Answer 16

Lol thanks lovely person! xD

Answer 17

http://dev.physicslab.org/Document.aspx?doctype=3&filename=UniversalGravitation_UniversalGravitationForces.xml This appears to be the target concept.

Answer 18

I dont get how they subbed dm=(M/L)dr

Answer 19

&Yes I shall read up.

Answer 20

@jennilalala what are you confused about?

Answer 21

I didnt understand how the solution manual defined dm as (M/L)dr

Answer 22

"Our first example involves calculating the gravitational force between a point mass M and an extended rod of mass m, length L, and mass per unit length, (lambda)"
a quote from the site that I think might answer your question

Answer 23

I don't see anything in the relevant example about radius

Answer 24

The site actually just confused me more so, throwing in lambda and whatnot.

Answer 25

ok, lambda is mass per length.

Answer 26

dm
___
dx

Answer 27

Oh derivative of..basically with the understanding that the mass of the rod particle being attracted I will substitute the dm/dr equation into my force of gravity one and simplify then integrate to get the magnitude force?

Answer 28

I think so, try it and see what you get.

Answer 29

Yeah I got the correct answer but I am still not sure why dm/dr=M/L... can you clarify that?

Answer 30

what do you mean by dr?

Answer 31

thats what my problem uses. r is the distance x

Answer 32

oh, I see.

Answer 33

dm/dx=m/L because the rod is of uniform composition

Answer 34

such a simple concept, and I go trying to create a formula proof for it xD lol

Answer 35

more funny is that it confused me until you asked XD lol