Calculating Frictional Force on Inclined Block | 43 kg Mass | 19° Inclination

AI Thread Summary
To calculate the frictional force on a 43 kg block resting on a 19° incline, the coefficients of static and kinetic friction are given as 0.42 and 0.36, respectively. The gravitational acceleration is 9.8 m/s². Since the block is at rest, the static friction is the relevant force to consider. By applying Newton's first law, one can identify the forces acting on the block, including gravitational and frictional forces, to determine the frictional force in Newtons. The solution involves calculating the normal force and then using the static friction coefficient to find the frictional force.
grouchy
Messages
72
Reaction score
0
a block is at rest on the incline shown in the figure. the coefficients of static and kinetic friction are s = 0.42 and k = 0.36 respectively. The acceleration of gravity is 9.8 m/s^2. the angle of inclination is 19 degrees.

what is the frictional force acting on the 43 kg mass? answer in units of N.


Honestly I have no idea what to do on this problem...
 
Physics news on Phys.org
grouchy said:
a block is at rest on the incline shown in the figure. the coefficients of static and kinetic friction are s = 0.42 and k = 0.36 respectively. The acceleration of gravity is 9.8 m/s^2. the angle of inclination is 19 degrees.

what is the frictional force acting on the 43 kg mass? answer in units of N.


Honestly I have no idea what to do on this problem...
If the block is at rest, you need only to identify all the forces acting on the block and apply Newton 1 to solve for the unknown friction force.
 
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