Forces on an inclined plane

  • Thread starter Koborl
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  • #1
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A Black Garden Ant is seen dragging a caterpillar with mass = 1.5g = .0015 KG
the ant manages to accelerate the caterpillar at 1mms^-2

Assume that friction between the caterpillar and the ground can be ignored.

(a) What force has the ant applied to cause this acceleration?


(b) The ant bites the caterpillar near to the ground and pulls it an angle of 30° to the horizontal.
What total force does it have to apply to move the caterpillar with the same horizontal acceleration?

(c) What is the normal force acting on the caterpillar whilst being pulled by the ant at 30°


have missed out on a week of classes due to being sick and have no idea what to do here.

EDIT: i think this is right now, but not sure


1.

a)
F = .0015kg x .001m/s
F = 0.0000015N

b)
let x = horizantal
let y = vertical
let z = hypotenenuse

cos 30 = x/z
f= x/cos 30
f = 1.732 x 10^-6

c) w = .015N
normal force(N):
N = w - y
N = .014999
 
Last edited:

Answers and Replies

  • #2
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Your drawing appears correct.

a) Your calculated force is wrong, you forgot to convert g to kg.
b) You need to find the force on the diagonal, use the Pythagorean theorem.
c) Normal force on a slope is calculated as mgcos[itex]\theta[/itex]
 
  • #3
tiny-tim
Science Advisor
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Welcome to PF!

Hi Koborl! Welcome to PF! :smile:
… have missed out on a week of classes due to being sick and have no idea what to do here.
hmm … you obviously missed all the information about black garden ants :redface:

nooo, your diagram is completely wrong, the ground is horizontal, the caterpillar is horizontal, but the ant is dragging it with a force at 30° to the horizontal (as if it was pulling it with a string at that angle)

in future, please go to your nearest black garden and study the ants there before attempting questions like this! :biggrin:

however does he expect to pass his physics exams without minoring in ant studies? :rolleyes:
 
  • #4
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Oh wow, totally didn't see the block ON the hill, thought he was pulling on it xD. Thanks tim.
 
  • #5
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1.[itex]Force(Kg \frac{m}{sec^2} or \frac {ML}{T^2})=ma \ N[/itex]

2. On inclined plane 2 forces, and the resultant is ma
The gravity that pull the caterpillar down, mgSin30°.
Thus F-mgSin30°=ma

3. Not sure N on level or inclined plane.

a) On level plane, the x component of force accelerate the caterpillar.
Fcos30°=ma
The vertical or Y component pulling the object up against the weight.
N=mg -FSin30°
b) On inclined plane
FCos30°-mgSin30°=ma
N=MgCos30°-FSin30°
[itex]N=mgCos30°-Sin30°(\frac {(ma+mgSin30°)}{Cos30°})[/itex]
 
Last edited:
  • #6
157
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1.[itex]Force(Kg \frac{m}{sec^2} or \frac {ML}{T^2})=ma \ N[/itex]

2. On inclined plane 2 forces, and the resultant is ma
The gravity that pull the caterpillar down, mgSin30°.
Thus F-mgSin30°=ma

3. Not sure N on level or inclined plane.

a) On level plane, the x component of force accelerate the caterpillar.
Fcos30°=ma
The vertical or Y component pulling the object up against the weight.
N=mg -FSin30°
b) On inclined plane
FCos30°-mgSin30°=ma
N=MgCos30°-FSin30°
[itex]N=mgCos30°-Sin30°(\frac {(ma+mgSin30°)}{Cos30°})[/itex]
N on inclined plane is mgcos[itex]\theta[/itex].
 

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