Acceleration Problem: Calculating Force of Pet 2

  • Thread starter Thread starter dimava
  • Start date Start date
  • Tags Tags
    Acceleration
AI Thread Summary
To solve the problem of calculating the force exerted by Pet 2 on the wagon, it is essential to analyze the forces acting on the wagon, which has a mass of 12 kg and rolls without friction. Pet 1 pulls at a force of 5 N at an angle of 15 degrees to the left, while Pet 2 pulls at 30 degrees to the right. The key is to find the X-components of the forces from both pets, ensuring they balance out for the wagon to accelerate straight forward. The formula derived indicates that the force exerted by Pet 2 can be calculated as x = (5*sin 15)/(sin 30). This approach confirms that the forces are equal and opposite, allowing for the desired acceleration of the wagon.
dimava
Messages
6
Reaction score
0
Hi guys,

I have a homework problem which I can use some help with:

Lisa harnesses two pets to a wagon. The mass of the wagon is 12 kg, and it rolls without friction on a horizontal floor. Pet 1 is pulling at 15 degrees left of forward with the force of 5 N. Pet 2 is pulling at 30 degrees right of forward. How much force does Pet 2 exert if the wagon acclerates straight forward?

I also attached a picture

thanks in advance,

Dimava
 

Attachments

  • image1.jpg
    image1.jpg
    3.6 KB · Views: 548
Physics news on Phys.org
Well, I'm kind of lazy to do actual math for you, but I will give you hints on doing it.

First off, when you got two pets moving in different directions like that, you will definitally want to find the tension in the ropes to the wagon. By finding the tension caused by the pulling, you should be able to find the combined force on the wagon.
 
I think I got it, does 2.588N sound about right?
 
I don't know where this attached picture is, but...

Code:
\<--^  ^-->/
 \15|  |30/
  \ |  | /
F=5\|_ |/F=x
   | \\ |
   | \\ |
   | \\ | <--- Wagon
    ----
So basically this is how it has to work. Since the wagon is going to accelerate straight forward, you want the X-Components of each force which pet 1 and pet 2 exert to be equal and opposite, and hence no acceleration in either the left or right direction.

the X-Component for pet 1 will be 5*sin 15. The X-Component for pet 2 must be equal so:

5*sin 15 = x sin 30

solving for x:

x := (5*sin 15)/(sin 30)

this will give you the force which pet 2 exerts on the wagon

Or at least this is how I interpreted the question.

edit:Ok, you posted heheh. Yes, that is the same answer as I got, so I'm assuming you did it correctly :)
 
Last edited:
I used almost the same thing, but did the cos of the supplements but same idea, thanks for your reply :smile:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top