Solving Friction Problem for 30.0 kg Packing Case in a 1500 kg Pickup Truck

  • Thread starter Thread starter Amar.alchemy
  • Start date Start date
  • Tags Tags
    Friction
AI Thread Summary
The discussion focuses on a physics problem involving a 30.0 kg packing case in a 1500 kg pickup truck, analyzing the friction forces during different accelerations. When the truck accelerates northward at 2.20 m/s², the friction force acts in the same direction as the truck's acceleration to prevent the case from sliding. Conversely, when the truck accelerates southward at 3.40 m/s², the friction force also acts northward, again aiding the case's movement with the truck. The key point is that friction always acts to oppose the relative motion between the truck and the packing case, ensuring the case moves with the truck. Understanding this relationship is crucial for solving similar physics problems involving friction and acceleration.
Amar.alchemy
Messages
78
Reaction score
0

Homework Statement


A 30.0-kg packing case is initially at rest on the floor of a 1500-kg pickup truck. The coefficient of static friction between the case and the truck floor is 0.30, and the coefficient of kinetic friction is 0.20. Before each acceleration given below, the truck is
traveling due north at constant speed. Find the magnitude and direction of the friction force acting on the case (a) when the truck accelerates at 2.20 m/s' northward and (b) when it accelerates at 3.40 m/s' southward.


Homework Equations


Newton's second law,static friction force and kinetic friction force equations

The Attempt at a Solution


I came to know that in both cases (ie a and b) frictional force is in the direction of acceleration of truck. Kindly explain me this??
 
Physics news on Phys.org
The force on the package from acceleration of the truck is south (when the truck is going north). This is in the frame of reference of the truck bed. The mass of the object resists the motion of the truck. The truck is the one that's going north, and it of course wants the package to go with it. If there was no friction the package would stay where it is as the truck drives out from beneath. So the truck is going to be exerting a force in whatever direction it accelerates, in order that the package would go with it. And it does this through the friction. Hence the force from acceleration in the frame of reference of the truck on the package is counter to the direction of the truck and the force of friction that is counter to this force is necessarily in the direction of acceleration.
 
Last edited:
Thanks LowlyPion :-)
 
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 '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