Force required to change the direction

[adjacent]

Homework Statement

I haven't studied forces acting on angles yet.
An object of mass 1kg is moving horizontally at a constant speed 10m/s.What is the the force required vertically to turn it 40 degrees to vertical?

Not a home work question

Homework Equations

The Attempt at a Solution

As I said,I haven't studied this yet.This is a homework like question.

 

[HallsofIvy]

Your question is not complete. Any amount of force, no matter how small will eventually change the direction by any desired angle. You need to specify a time interval.

(The force acting vertically is not enough. In order that there be no change in speed, the force must act perpendicular to the velocity vector so the force must change direction as the object does.)

(Surely you are not thinking that the direction will change instantaneously[/b] by 40 degrees? The path of the path of the object will curve to that direction.)

 

[adjacent]

I see.Change in speed is not a matter.What's the equation relating force,time and angle?

 

[nasu]

I am not sure what you mean by the above (change in speed is not a matter).
If you assume constant speed, you could calculate the change in momentum between the two states: moving horizontally and moving at an angle of 40 degrees.
This change in momentum will give you the impulse of the force:
Δmv=F*Δt
where v and F are vectors and F is the average force.

 

[Enigman]

I see.Change in speed is not a matter.What's the equation relating force,time and angle?

Its better to think in terms of momentum.
F=dp/dt
F=dmv/dt (well, classically speaking...)
F=ma
Now to find a relation between F, t, θ and p.
F-force t-time θ-angle p-momentum
Take the initial direction of p (ie. p(0)) along x-axis ($\hat{i}$))
Let force act in the x-y plane and $\hat{i}$ be along x axis.
F= a$\hat{i}$+b$\hat{j}$
Now at any time t-
p(t)=at+p$\hat{i}$+bt$\hat{j}$
∴tan(θ)=bt/(at+p)
θ=tan^-1(bt/(at+p))
But equations like these are never used or necessary and usually derived to suit the needs.

 

[adjacent]

Its better to think in terms of momentum.
F=dp/dt
F=dmv/dt (well, classically speaking...)
F=ma
Now to find a relation between F, t, θ and p.
F-force t-time θ-angle p-momentum
Take the initial direction of p (ie. p(0)) along x-axis ($\hat{i}$))
Let force act in the x-y plane and $\hat{i}$ be along x axis.
F= a$\hat{i}$+b$\hat{j}$
Now at any time t-
p(t)=at+p$\hat{i}$+bt$\hat{j}$
∴tan(θ)=bt/(at+p)
θ=tan^-1(bt/(at+p))
But equations like these are never used or necessary and usually derived to suit the needs.

What is a and b?

 

[Enigman]

Er...components of the Force vector. a is the force along x-axis or the unit vector $\hat{i}$ and b is force along y-axis or along $\hat{j}$. When a$\hat{i}$ is added to b$\hat{j}$ we get the net force.
If you haven't done vectors yet- http://www.dummies.com/how-to/content/how-to-find-vector-components.html

 

[adjacent]

There is no force on the x axis.It's moving with a constant velocity.A force is applied on the y axis

 

[Enigman]

There is no force on the x axis.It's moving with a constant velocity.A force is applied on the y axis

Then a=0.
tan(θ)=bt/(p)
At θ=90° tanθ does not exist, therefore it can never turn to 90°
This happens because as there is no force along x-axis the momentum along it does not change. Then it follows that there is always some velocity acting along x axis.