What is the Required Horizontal Force for a Shopping Cart on an Inclined Plane?

  • Thread starter Thread starter whoopie88
  • Start date Start date
  • Tags Tags
    Forces Incline
AI Thread Summary
To determine the required horizontal force to push a 7.3 kg shopping cart up a 13° incline with an acceleration of 1.67 m/s², one must consider the forces acting on the cart. The applied force is horizontal, which complicates the calculation as it differs from the incline's angle. After analyzing the forces, the correct approach leads to the solution for the horizontal force needed. The discussion highlights the importance of understanding force components in inclined scenarios. Ultimately, the problem is resolved with the correct application of physics principles.
whoopie88
Messages
52
Reaction score
0

Homework Statement


A shopper pushes a 7.3 kg shopping cart up a 13° incline, as shown in Figure 5-24. Find the magnitude of the horizontal force, F, needed to give the cart an acceleration of 1.67 m/s2.

05-21.gif

Figure 5-24

Homework Equations


All forces equations.


The Attempt at a Solution


wrhoas.jpg


Help?
 
Physics news on Phys.org
Realize that the applied force is horizontal, not parallel to the incline.
 
Ohhh, thank you, I got the answer!
 
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