Acceleration and for direction

[junkie_ball]

Hi,

I've just started studying a mechanical engineering course and have a real newbie question. If i have a force acting on a object at an angle to it above the horizontal that will be providing the velocity but the object will be moving along the horizontal to work out the acceleration will i need to calculate it using the force acting on it at the angle or the resultant horizontal force?

I hope that make sense?

 

[tiny-tim]

welcome to pf!

hi junkie_ball! welcome to pf!

If i have a force acting on a object at an angle to it above the horizontal that will be providing the velocity but the object will be moving along the horizontal to work out the acceleration will i need to calculate it using the force acting on it at the angle or the resultant horizontal force?

you use the horizontal component (not "resultant" … a resultant is the force that results from adding two or more forces … though you can say "resolved") of the force

the reason is that everything has to be in the same direction …

the velocity (and the acceleration) is purely horizontal, so you resolve everything into horizontal components

 

[Doc Al]

What determines the acceleration is the net force on the object. If the object only moves horizontally, you know that the net force has no vertical component. So all you need worry about is the horizontal component of the force.

 

[junkie_ball]

Thanks for the replies. So if i get this correctly i need to work the acceleration out on the horizontal. I have attached a diagram of the question. I'm not looking for an overall answer but clarification of my thinking.

In the attached the crate will be accelerated using a dragline at an angle but only move along the rollers horizontally. As such i will need to calculate the acceleration using the horizontal force which i can work out with trig. Is that correct?

 

[tiny-tim]

hi junkie_ball!

In the attached the crate will be accelerated using a dragline at an angle but only move along the rollers horizontally.

As such i will need to calculate the acceleration using the horizontal force which i can work out with trig. Is that correct?

yes, that's correct

(and yes, the horizontal component is the force times cos of the angle between)

 

[junkie_ball]

hi junkie_ball!


yes, that's correct

(and yes, the horizontal component is the force times cos of the angle between)

Thanks for the confirmation!

 

[junkie_ball]

Hi thanks for all you previous replies most helpful in my understanding. I now have one further query on the problem i attached in my previous post. To calculate work done i understand the equation is:

Work Done = Force x Displacement

As my displacement is in the horizontal plane does this mean i calculate the WD by using the horizontal force which i have worked out using trig or do i need to use the 50N force connecting the crate and motor? I'm assuming it's the horizontal force?

 

[Doc Al]

I'm assuming it's the horizontal force?

Yes, you need to use the component of force in the direction of the displacement, in this case horizontal.

 

[tiny-tim]

if you understand the scalar product (dot product), it's actually

Work Done = Force "dot" Displacement ​