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# M1 inclined planes Q Watch

1. I was going through a past paper and came across this:

A small package of mass 1.1 kg is held in equilibrium on a rough plane by a horizontal force. The plane is inclined at an angle α to the horizontal, where tan α=3/4. The force acts in a vertical plane containing a line of greatest slope of the plane and has magnitude P newtons, as shown in Figure 2. The coefficient of friction between the package and the plane is 0.5 and the package is modelled as a particle. The package is in equilibrium and on the point of slipping down the plane.

(a) Draw, on Figure 2, all the forces acting on the package, showing their directions clearly.

(2)(b) (i) Find the magnitude of the normal reaction between the package and the plane.

For Q 2.b.i, on the mark scheme, it said to use the formula R(↑), Rcosα + Fsinα = mg. How would I come to this conclusion, as my first instinct was to have R = 1.1gCosα ?
2. they are probably showing you what it looks like before you put the digits in, and how many forces did you draw on i put 4 excluding the weight, but it does not work out, there must be an extra force acting on it, maybe acceleration? or probably another thing.
3. (Original post by najinaji)
I was going through a past paper and came across this:

A small package of mass 1.1 kg is held in equilibrium on a rough plane by a horizontal force. The plane is inclined at an angle α to the horizontal, where tan α=3/4. The force acts in a vertical plane containing a line of greatest slope of the plane and has magnitude P newtons, as shown in Figure 2. The coefficient of friction between the package and the plane is 0.5 and the package is modelled as a particle. The package is in equilibrium and on the point of slipping down the plane.

(a) Draw, on Figure 2, all the forces acting on the package, showing their directions clearly.

(2)(b) (i) Find the magnitude of the normal reaction between the package and the plane.

For Q 2.b.i, on the mark scheme, it said to use the formula R(↑), Rcosα + Fsinα = mg. How would I come to this conclusion, as my first instinct was to have R = 1.1gCosα ?
When you draw the diagram, you will see the forces acting. They are resolving vertically, so the component of friction will also need to be considered.

The way you are trying to resolve is perpendicular to the plane, but your equation is wrong according to the question, because you will need to consider the component of 'P' as well.

So if you want to resolve perpendicular to the plane, then the equation will be,

Though i will advice you to always try to resolve perpendicular to the plane, because if it isn't in equilibrium then you will need to consider the acceleration component as well.
4. (Original post by Maphs)
they are probably showing you what it looks like before you put the digits in, and how many forces did you draw on i put 4 excluding the weight, but it does not work out, there must be an extra force acting on it, maybe acceleration? or probably another thing.
The mark scheme equation is correct, the question says that the particle is in equilibrium, hence acceleration is zero.
5. (Original post by raheem94)
When you draw the diagram, you will see the forces acting. They are resolving vertically, so the component of friction will also need to be considered.

The way you are trying to resolve is perpendicular to the plane, but your equation is wrong according to the question, because you will need to consider the component of 'P' as well.

So if you want to resolve perpendicular to the plane, then the equation will be,

Though i will advice you to always try to resolve perpendicular to the plane, because if it isn't in equilibrium then you will need to consider the acceleration component as well.
Would there have been any easy way to find the answer if I resolved the way you described?
6. (Original post by najinaji)
Would there have been any easy way to find the answer if I resolved the way you described?
I don't see anything difficult in both ways. It just requires practice, to become good at such stuff.

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