A 4.4kg block is pushed along the ceiling

  • Thread starter Thread starter madelines
  • Start date Start date
  • Tags Tags
    Block
AI Thread Summary
A 4.4 kg block is pushed along the ceiling with a constant force of 84.0 N at a 50° angle, resulting in an acceleration of 6.92 m/s². The calculation for the coefficient of kinetic friction (u_k) yielded a value greater than 1, which raised concerns about its validity. It was clarified that while coefficients of friction are typically less than 1, values above 1 can occur in certain materials, indicating that the force required to move the object exceeds the normal force. The calculations for normal force (F_n) and frictional force (F_k) were also discussed, emphasizing the importance of accurate computations. This discussion highlights the nuances of friction coefficients in physics problems.
madelines
Messages
2
Reaction score
0

Homework Statement



A 4.40 kg block is pushed along the ceiling with an constant applied force of F = 84.0 N that acts at an angle θ = 50° with the horizontal. The block accelerates to the right at 6.92 m/s2. Determine the coefficient of kinetic friction between the block and the ceiling.


Homework Equations


u_k=(F_k)/(F_n)
ƩF=m*a
(pythagorean theorem for F_n)


The Attempt at a Solution


I've tried this problem multiple different ways, but I keep getting a number that is greater than 1. Thank you!

F_n=(4.4*9.8)-(84*sin50)=-21.227733
-F_k=(4.4*6.92)-(84*cos50)=-23.546259
u_k=1.1092
but 0<u_k<1
 
Physics news on Phys.org
I haven't checked your working but..

http://en.wikipedia.org/wiki/Friction#Coefficient_of_friction

Occasionally it is maintained that µ is always < 1, but this is not true. While in most relevant applications µ < 1, a value above 1 merely implies that the force required to slide an object along the surface is greater than the normal force of the surface on the object. For example, silicone rubber or acrylic rubber-coated surfaces have a coefficient of friction that can be substantially larger than 1.
 
Thank you! That cleared up some of the confusion!
 
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