How Does Newton's Third Law Apply to a Particle on Different Slopes?

[ecy5maa]

Homework Statement

A particle of mass 4kg is being towed at a constant speed up a rough plane inclined at 30 degrees to the horizontal by a force 4g N acting parallel to the slope. At the top of the slope the particle moves onto a rough horizontal slope with the same coefficient of friction. If the towing force continues to act in the same direction, show that the particle undergoes an acceleration of

$$\frac{g\sqrt{3}}{6}$$ m.second square

Homework Equations

basically use F=ma and get the answer where m=4kg

The Attempt at a Solution

1. On the inclined plane coefficient of friction was calculated as 1/ (sqrt 3)

2.Since they say the force of 4g will be acting in the same direction even when it is on the horizontal plane i took this too mean that 4g will be acting at an angle 30 degrees to the horizontal when the particle is on a horizontal plane..

This way Reaction force would be 4g-4gsin30= 2g

And net force would be equal too

4a= 4gcos30 - 2g/ (sqrt 3)


which makes a = $$\frac{g\sqrt{3}}{3}$$ m.second square which is twice the answer that is given.


Can some one please check this out. Its from an A level Mechanics book

 

[collinsmark]

For what it's worth, I also ended up with your answer that $$a = g\sqrt{3}/3$$

 

[SammyS]

Homework Statement

A particle of mass 4kg is being towed at a constant speed up a rough plane inclined at 30 degrees to the horizontal by a force 4g N acting parallel to the slope. At the top of the slope the particle moves onto a rough horizontal slope with the same coefficient of friction. If the towing force continues to act in the same direction, show that the particle undergoes an acceleration of

$$\frac{g\sqrt{3}}{6}$$ m.second square

Homework Equations

basically use F=ma and get the answer where m=4kg

The Attempt at a Solution

1. On the inclined plane coefficient of friction was calculated as 1/ (sqrt 3)


Can some one please check this out. Its from an A level Mechanics book

Show how you calculate the coefficient of friction, μ.

I get, $$\mu=\frac{2}{\sqrt{3}}\,.$$

Added in an edit. Ignore this post!

 

[collinsmark]

Show how you calculate the coefficient of friction, μ.

I get, $$\mu=\frac{2}{\sqrt{3}}\,.$$

I ended up with μ = 1/√3, which agrees with the original poster. (The pulling force must be equal in magnitude to the frictional force plus the component of gravitational force parallel to the slope.) But I'll let ecy5maa comment further.

 

[SammyS]

I ended up with μ = 1/√3, which agrees with the original poster. (The pulling force must be equal in magnitude to the frictional force plus the component of gravitational force parallel to the slope.) But I'll let ecy5maa comment further.

I agree.

μ = 1/√3