# Friction Question Help

"A block of mass 5kg rests on a rough plane inclined at 27° to the horizontal. A force of 35N acts parallel to the plane, and the block is on the point of moving up the plane. Find:

(a) The reaction force between the block and the plane (3 s.f.)

(b) The coefficient of friction between the block and the plane (3 s.f.)"

I've managed to do part a by using R - 5gcos(27°) = 0 and then rearranging to find R, but I can't figure out how to solve for part b, I've tried using the 35 given in the equation then dividing that by the Reaction force (using equation f = uR) to get the coefficient, this gave me 0.802, which is wrong according to Integral Maths, and I'm not particularly sure what else to do.

I've attached my working out as well as a picture of the diagram I have drawn for the question, any help would be appreciated.
Original post by Arecedia
"A block of mass 5kg rests on a rough plane inclined at 27° to the horizontal. A force of 35N acts parallel to the plane, and the block is on the point of moving up the plane. Find:

(a) The reaction force between the block and the plane (3 s.f.)

(b) The coefficient of friction between the block and the plane (3 s.f.)"

I've managed to do part a by using R - 5gcos(27°) = 0 and then rearranging to find R, but I can't figure out how to solve for part b, I've tried using the 35 given in the equation then dividing that by the Reaction force (using equation f = uR) to get the coefficient, this gave me 0.802, which is wrong according to Integral Maths, and I'm not particularly sure what else to do.

I've attached my working out as well as a picture of the diagram I have drawn for the question, any help would be appreciated.

You resolved perpendicular to the plane to determine R. Resolve along the plane to find mu. You have 3 forces to balance: 35, friction and resolved weight.
(edited 10 months ago)
Original post by Arecedia
"A block of mass 5kg rests on a rough plane inclined at 27° to the horizontal. A force of 35N acts parallel to the plane, and the block is on the point of moving up the plane. Find:

(a) The reaction force between the block and the plane (3 s.f.)

(b) The coefficient of friction between the block and the plane (3 s.f.)"

I've managed to do part a by using R - 5gcos(27°) = 0 and then rearranging to find R, but I can't figure out how to solve for part b, I've tried using the 35 given in the equation then dividing that by the Reaction force (using equation f = uR) to get the coefficient, this gave me 0.802, which is wrong according to Integral Maths, and I'm not particularly sure what else to do.

I've attached my working out as well as a picture of the diagram I have drawn for the question, any help would be appreciated.

F = uR relates the frictional force to the normal reaction on the point of motion, the "F" is not your 35N applied force.
Original post by Arecedia
"A block of mass 5kg rests on a rough plane inclined at 27° to the horizontal. A force of 35N acts parallel to the plane, and the block is on the point of moving up the plane. Find:

(a) The reaction force between the block and the plane (3 s.f.)

(b) The coefficient of friction between the block and the plane (3 s.f.)"

I've managed to do part a by using R - 5gcos(27°) = 0 and then rearranging to find R, but I can't figure out how to solve for part b, I've tried using the 35 given in the equation then dividing that by the Reaction force (using equation f = uR) to get the coefficient, this gave me 0.802, which is wrong according to Integral Maths, and I'm not particularly sure what else to do.

I've attached my working out as well as a picture of the diagram I have drawn for the question, any help would be appreciated.

I ******* hate integral maths
Original post by Arecedia
"A block of mass 5kg rests on a rough plane inclined at 27° to the horizontal. A force of 35N acts parallel to the plane, and the block is on the point of moving up the plane. Find:

(a) The reaction force between the block and the plane (3 s.f.)

(b) The coefficient of friction between the block and the plane (3 s.f.)"

I've managed to do part a by using R - 5gcos(27°) = 0 and then rearranging to find R, but I can't figure out how to solve for part b, I've tried using the 35 given in the equation then dividing that by the Reaction force (using equation f = uR) to get the coefficient, this gave me 0.802, which is wrong according to Integral Maths, and I'm not particularly sure what else to do.

I've attached my working out as well as a picture of the diagram I have drawn for the question, any help would be appreciated.

You forgot the horizontal compoment of 5g
You need to use 35-5gsin27-F=0 now as F = UR sub R so 35-5gSin(27)-U(5gcos(27)=0
U(cos(27))=35-5gsin(27)
and i think U= 35-5gsin(27) divided by cos(27) is the answer. i might of made a stupid mistake but that is the idea
Original post by ffffffffffffff.
You forgot the horizontal compoment of 5g
You need to use 35-5gsin27-F=0 now as F = UR sub R so 35-5gSin(27)-U(5gcos(27)=0
U(cos(27))=35-5gsin(27)
and i think U= 35-5gsin(27) divided by cos(27) is the answer. i might of made a stupid mistake but that is the idea

Between your second and third lines of text, (mu)5gcos(27) has (wrongly) become (mu)cos(27).

Note also that this thread is seven months old.