Simple force question that I cant get

[mikejones2000]

I know I got at least two out of the three incorrect and cannot seem to understand why.
The question states:
The coefficients of friction between a block of mass 1 kg and a surface are μs=0.64 and μk=0.42. Assume the only other force acting on the block is that due to gravity, what is the magnitude of the frictional force on the block


A)if the block is at rest and the surface is horizontal?
B)if the block is at rest and the surface is inclined at 22o?
C)if the block is at moving downward and the surface is inclined at 49o?

For a) I said the magnitude is 0 because there is no force acting on it other than gravity and for b) I put .64-9.8sin(22) and c).42-9.8sin49. I got two negative numbers for b) and c) and probably think those are the ones I screwed up on but cannot understand how if the only other force acting on it is gravity and not a normal force. Any help would be greatly appreciated.

 

[Doc Al]

I think you took them too literally when they said "Assume the only other force acting on the block is that due to gravity". Of course there's a normal force, otherwise the frictional force will always be zero. They meant no other applied forces, like someone pushing or pulling the block.

 

[kyle8921]

The magnitude of the frictional force on the block can be calculated using the formula Ff = μN, where μ is the coefficient of friction and N is the normal force. In this case, the normal force is equal to the weight of the block, which is mg, where m is the mass of the block and g is the acceleration due to gravity.

For part A, the block is at rest on a horizontal surface, so the normal force is equal to the weight of the block, which is 1 kg * 9.8 m/s^2 = 9.8 N. Therefore, the magnitude of the frictional force is Ff = μN = 0.64 * 9.8 = 6.272 N.

For part B, the block is at rest on an inclined surface, so the normal force is equal to the component of the weight of the block perpendicular to the surface, which is mgcosθ, where θ is the angle of inclination. In this case, θ = 22o, so the normal force is 1 kg * 9.8 m/s^2 * cos22o = 9.327 N. Therefore, the magnitude of the frictional force is Ff = μN = 0.64 * 9.327 = 5.973 N.

For part C, the block is moving downward on an inclined surface, so the normal force is equal to the component of the weight of the block perpendicular to the surface, which is mgcosθ, where θ is the angle of inclination. In this case, θ = 49o, so the normal force is 1 kg * 9.8 m/s^2 * cos49o = 6.154 N. Therefore, the magnitude of the frictional force is Ff = μN = 0.42 * 6.154 = 2.585 N.

It is important to note that the normal force changes depending on the orientation of the surface, and therefore the magnitude of the frictional force also changes. This is why the calculations for parts B and C are different from part A. Additionally, the frictional force always acts in the opposite direction of the motion, which is why the values for parts B and C are negative.