2 dimension forces - problem solving for parked car

AI Thread Summary
To determine if the parked car will remain at rest, the gravitational force acting on the car must be resolved into its perpendicular and parallel components relative to the incline. The weight of the car is 20,000N, and the parallel component of this force can be calculated using the sine of the incline angle (30 degrees). This results in a parallel force of approximately 10,000N, which is less than the maximum brake force of 12,000N. Therefore, the car will remain at rest since the braking force exceeds the force trying to move it down the incline. Understanding how to resolve forces into components is crucial for solving such problems.
maryemilyw
Messages
1
Reaction score
0
Problem of the day: Trying to find the formula to use for my 15 year old daughter.
A 20,000N car is parked on an incline that makes an angle of 30 degrees with the horizontal. If the maximum force the brakes can withstand is 12,000N, will the car remain at rest?
She is just learning how to pick formula and plug in the proper info.

HELP!
 
Physics news on Phys.org
Just resolve the vector into it's perpendicular & parallel components. Have you done this before?
 
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