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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 to the horizontal, where tan[a] = 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.

^^ i know how to do (a)

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

(ii) Find the value of P.

im lost here ;o

i thought ur meant to resolve parallel and perpendicular but the markscheme they resolve horizontally and vertically ;o

can anyone explain how they got this

Rcosα + Fsinα = mg

R = 1.1g /(cosα + [1/2]sinα ) = 9.8 N

and P + [1/2]Rcosα = Rsinα

P = R(sinα − [1/2]cosα ) = 1.96

thxs

r-dragon2k8

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 to the horizontal, where tan[a] = 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.

^^ i know how to do (a)

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

(ii) Find the value of P.

im lost here ;o

i thought ur meant to resolve parallel and perpendicular but the markscheme they resolve horizontally and vertically ;o

can anyone explain how they got this

Rcosα + Fsinα = mg

R = 1.1g /(cosα + [1/2]sinα ) = 9.8 N

and P + [1/2]Rcosα = Rsinα

P = R(sinα − [1/2]cosα ) = 1.96

thxs

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 to the horizontal, where tan[a] = 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.

^^ i know how to do (a)

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

(ii) Find the value of P.

im lost here ;o

i thought ur meant to resolve parallel and perpendicular but the markscheme they resolve horizontally and vertically ;o

can anyone explain how they got this

Rcosα + Fsinα = mg

R = 1.1g /(cosα + [1/2]sinα ) = 9.8 N

and P + [1/2]Rcosα = Rsinα

P = R(sinα − [1/2]cosα ) = 1.96

thxs

b. i)

Resolving the forces acting perpendicular to the plane, acting on the box, we have; (using f=ma on the box, taking upwards perpendicular to the plane as positive)

R+Psina-mgcosa=0 , acceleration is 0 as not moving, m is the mass of the box, a is the angle of the plane with the horizontal. We know everything other than R, the normal reaction, so rearrange to calculate it. This gives the forces acting on the box perpendicular to the plane.

Parallel to the plane we have; (taking down the plane as positive)

-Pcosa-(mu)R=0 , where mu is the coefficient of friction. Using these two equations, we can solve them simultaneously to get values for R and P

r-dragon2k8

im lost here ;o

i thought ur meant to resolve parallel and perpendicular but the markscheme they resolve horizontally and vertically ;o

can anyone explain how they got this

Rcosα + Fsinα = mg

R = 1.1g /(cosα + [1/2]sinα ) = 9.8 N

and P + [1/2]Rcosα = Rsinα

P = R(sinα − [1/2]cosα ) = 1.96

thxs

im lost here ;o

i thought ur meant to resolve parallel and perpendicular but the markscheme they resolve horizontally and vertically ;o

can anyone explain how they got this

Rcosα + Fsinα = mg

R = 1.1g /(cosα + [1/2]sinα ) = 9.8 N

and P + [1/2]Rcosα = Rsinα

P = R(sinα − [1/2]cosα ) = 1.96

thxs

The usual method is to resolved parallel and perpendicular to the plane, BUT there is no requirement to do it in those two directions. You could resolve in almost any directions you choose.

However, most directions you choose would lead to messy mathematics, with having to apply the sine or cosine to every force to get its value in the particular direction.

We choose directions to make life easier, and that means at right angles to each other, and parallel/perpendicular to some of the forces.

You could work this problem by resolving parallel and perpendicular to the plane, and there would be nothing wrong with that.

Since we have a horizontal force and a vertical force, it is as easy to resolve vertically and horizontally.

There is one BIG CAVEAT. The mass is in equilibrium. If the mass was accelerating down the slope, then you would resolve parallel and perpendicular to the slope, end of story. To use any other directions would be somewhat masochistic (but would still lead to the correct answers).

As to the actual values of the quoted equations, have a go yourself and post your working if you get stuck. The principle behind them of resolving a force in a given direction is the same one you use when you resolve parallel and perpendicular to the plane.

when i see these diagrams the first thing that i think of is :

F=MA resultant force = mass x acceleration

and

U=F/R Coefficient of friction = Friction / Normal reaction force

firstly i use the weight of the partical to calculate the resultant force perpandicular to the slope so i can work out the friction.

secondly i resolve the forces parallel to the slope and subtract friction away from them to find the resultant force.

then substitute in F=MA to find acelleration

^^^^^ thats just my method for these sort of questions ^^^^^

but for yours you have P so you need to resolve it as an unknown then do some simultaneous equations

F=MA resultant force = mass x acceleration

and

U=F/R Coefficient of friction = Friction / Normal reaction force

firstly i use the weight of the partical to calculate the resultant force perpandicular to the slope so i can work out the friction.

secondly i resolve the forces parallel to the slope and subtract friction away from them to find the resultant force.

then substitute in F=MA to find acelleration

^^^^^ thats just my method for these sort of questions ^^^^^

but for yours you have P so you need to resolve it as an unknown then do some simultaneous equations

Original post by r-dragon2k8

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 to the horizontal, where tan[a] = 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.

^^ i know how to do (a)

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

(ii) Find the value of P.

im lost here ;o

i thought ur meant to resolve parallel and perpendicular but the markscheme they resolve horizontally and vertically ;o

can anyone explain how they got this

Rcosα + Fsinα = mg

R = 1.1g /(cosα + [1/2]sinα ) = 9.8 N

and P + [1/2]Rcosα = Rsinα

P = R(sinα − [1/2]cosα ) = 1.96

thxs

As ghostwalker says, you can resolve forces in any two mutually perpendicular directions. For equilibrium you can do horizontal and vertical, or parallel and perpendicular.

If you have motion down the plane then it is natural to resolve parallel and perpendicular since motion takes place in the "parallel" direction and you can basically use F = ma to calculate the acceleration directly if you have a force in that direction.

Original post by davros

As ghostwalker says, you can resolve forces in any two mutually perpendicular directions. For equilibrium you can do horizontal and vertical, or parallel and perpendicular.

If you have motion down the plane then it is natural to resolve parallel and perpendicular since motion takes place in the "parallel" direction and you can basically use F = ma to calculate the acceleration directly if you have a force in that direction.

If you have motion down the plane then it is natural to resolve parallel and perpendicular since motion takes place in the "parallel" direction and you can basically use F = ma to calculate the acceleration directly if you have a force in that direction.

Not sure if you have realised how long ago this was

Original post by Super199

Not sure if you have realised how long ago this was

Oops - caught out by another pointless thread resurrection

Original post by davros

Oops - caught out by another pointless thread resurrection

resuurect!!!

but i dont get how u know what f is

- Unit 19 health and social care, HELP PLEASE!!
- Numbers of Arrangements
- the issac physics astronomy question
- A level chemistry help please
- btec science unit 7 part a release 2023 january
- A Chain Sliding Through a Tube
- Weighing at Sea
- Edexcel IAL M1 May 2024
- What units do I need for Edexcel IAL or IAS Further Maths
- Can i give IAL Edexcel Mathmematics across 3 sessions?
- M1 suvat help pls my exam is this tuesday!!
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- Solomon or Elmwood or Delphis Papers?
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can someone please explain what principle domain is and why the answer is a not c?Maths

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