frictional forces?

Question

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 is 19 degrees.

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

Answer 1

First ID the physics involved. Here we are looking at interacting forces; so we are talking about Newton's laws.

Now ID all the forces (since we know this is a force problem).

Weight W = mg is always a force when there is mass = 43 kg in this case. g = 9.8 m/sec^2; so you can find the weight. But which direction does W act? As a force vector we need to know both its magnitude and its direction.

Friction is the source of another force F = sN; where s = .42 for static friction and N = W cos(theta), the normal weight pressing into the ramp at theta = 19 degrees incline. [Can you say why s and not k is used?]

Now turn to Newton...f = ma; where f is the net force acting along the ramp surface = W sin(theta) - F = 0. f = 0 because the block is static, there is no acceleration; so a = 0 and ma = 0 = f. W sin(theta) is that portion of the block's weight that is pointed downward along (parallel to) the surface of the inclined ramp. [You can see this by setting theta = 0 so there is no incline and the weight along the surface is zero; that is, all the weight acts perpendicular to the surface of the ramp.]

Thus f = ma = 0 = W sin(theta) - F; so that F = W sin(theta) = mg sin(theta) and we see the friction force F is the equal, but opposite force to the weight along the ramp. You can do the math.

Answer 2

43kg *9.8 m/s^2= 421N... its the weight of the block, as well as the resultant force... the Y component of the force is 421.4sin(90-19)=416.21N

for static friction you times it by 0.42
416.21*0.42=174.81 N

for knetic friction you times it by 0.36
416.21*0.36=149.84 N

Answer 3

I dont see the picture, However you can use this formula
FK = μK mg Cos 19