In This Question Use g=9.81ms

A particle of mass 4 kg rests on a rough horizontal surface in limiting equilibrium.

The coefficient of friction between the particle and the plane is 0.35.

A pulling force of P newtons is applied to the particle inclined at 15° to the horizontal.

Find the normal contact force between the particle and the surface in terms of P.

[2 marks]

So I’ve resolved P such that a force of Psin(15) is going upwards, while Pcos(15) is going to the right

Using F=ma, I’ve created 2 simultaneous equations & then got an answer of 35.84 + Psin(15)

This doesn’t look correct. Does anyone know what the answer is?

A particle of mass 4 kg rests on a rough horizontal surface in limiting equilibrium.

The coefficient of friction between the particle and the plane is 0.35.

A pulling force of P newtons is applied to the particle inclined at 15° to the horizontal.

Find the normal contact force between the particle and the surface in terms of P.

[2 marks]

So I’ve resolved P such that a force of Psin(15) is going upwards, while Pcos(15) is going to the right

Using F=ma, I’ve created 2 simultaneous equations & then got an answer of 35.84 + Psin(15)

This doesn’t look correct. Does anyone know what the answer is?

Original post by Sam.T.

In This Question Use g=9.81ms

A particle of mass 4 kg rests on a rough horizontal surface in limiting equilibrium.

The coefficient of friction between the particle and the plane is 0.35.

A pulling force of P newtons is applied to the particle inclined at 15° to the horizontal.

Find the normal contact force between the particle and the surface in terms of P.

[2 marks]

So I’ve resolved P such that a force of Psin(15) is going upwards, while Pcos(15) is going to the right

Using F=ma, I’ve created 2 simultaneous equations & then got an answer of 35.84 + Psin(15)

This doesn’t look correct. Does anyone know what the answer is?

A particle of mass 4 kg rests on a rough horizontal surface in limiting equilibrium.

The coefficient of friction between the particle and the plane is 0.35.

A pulling force of P newtons is applied to the particle inclined at 15° to the horizontal.

Find the normal contact force between the particle and the surface in terms of P.

[2 marks]

So I’ve resolved P such that a force of Psin(15) is going upwards, while Pcos(15) is going to the right

Using F=ma, I’ve created 2 simultaneous equations & then got an answer of 35.84 + Psin(15)

This doesn’t look correct. Does anyone know what the answer is?

Why doesn't it look correct? You made a sign error and it should be -Psin(15) instead.

Vertical forces balance means $R+P\sin(15) = mg$. So I am not sure where you got 35.84 from exactly, but it's close so what it should be.

No need to solve simultaneous equations here at all.

Original post by RDKGames

Why doesn't it look correct? You made a sign error and it should be -Psin(15) instead.

Vertical forces balance means $R+P\sin(15) = mg$. So I am not sure where you got 35.84 from exactly, but it's close so what it should be.

No need to solve simultaneous equations here at all.

Vertical forces balance means $R+P\sin(15) = mg$. So I am not sure where you got 35.84 from exactly, but it's close so what it should be.

No need to solve simultaneous equations here at all.

Thank You

So to double check since R + Psin(15) = 4g

then R = 4g - Psin(15)

So R = 39.24 - Psin(15) (Using g = 9.81)

Original post by Sam.T.

Thank You

So to double check since R + Psin(15) = 4g

then R = 4g - Psin(15)

So R = 39.24 - Psin(15) (Using g = 9.81)

So to double check since R + Psin(15) = 4g

then R = 4g - Psin(15)

So R = 39.24 - Psin(15) (Using g = 9.81)

Yep, nothing complicated, it's just 2 marks after all.

Original post by RDKGames

Yep, nothing complicated, it's just 2 marks after all.

Yeah I’ve think I over complicated it

Thanks a lot

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