What is the direction of the third force in my physics homework problem?

[ChinToka]

I got problems with with some of my homework problems in physics. Its about Newtons Laws of motion and i have really no idea what to do

I have an object beind acted upon by three forces and moves with a constant velocity. One force is 60N along the x-axis, the second is 75N along a direction making a counterclockwise angle of 150° with the x-axis. What I need to find out is the direction of the third force, measured conterclockwise from the x-axis.

I am not very good at vectors and trigonometry, so if someone can give me some advice where to start and with what formula, I would very appreciate it.

 

[ZapperZ]

Please note that we DO have a homework help section.

Zz.

 

[Nylex]

Draw a diagram to start, as they can help (even if they're quite simple). What condition must the forces on the object satisfy for it to be moving at a constant velocity?

 

[Dr.Brain]

As the object moves with constant velocity, its acceleration is zero , therefore this implies that net force on this object is zero.Place the particle at origin, and display the forces in directions as described in your post . Make use of simple geometry and methods to find resultant between vectors when the angle between them is given . Find the condition when net force on object is zero.

BJ

 

[professor]

well I am pretty sure what your looking for is a blaance from the counter clockwise angle... in such a case, the third force would be 75N acting in a direction that would create a clockwise angle of 150 degerrs with the x axis... that would put it back to a constant stright velocity along the x axis

 

[ChinToka]

@ZapperZ
oops, sry first post

@Nylex
I did, but i guess it was wrong or not helpful

What I need to find out is the angle of the third force counterclockwise to the x-axis

 

[Doc Al]

What I need to find out is the angle of the third force counterclockwise to the x-axis

Add the two forces (vectors) that you are given. The third force must be equal and opposite to that sum.

Can you find the x & y components of the forces?

 

[ChinToka]

now while your asking... i don´t know. Adding vectors is done by squareroot( x^2+y^2) and I guess the components are 60N and 75N.Squareroot(3600+5625) equals squareroot(9225) equals 96.04. Then 360° - 96 because it is counterclockwise which gives me 263°.
But I think its wrong.

BTW I am doing an independent study of "General Physics I" college, that´s why i have no clue of anything

 

[Doc Al]

Yes, it's wrong. The first step is to find the x & y components of each force. Given the angle to the x-axis, the components of a vector (A) are:
A_x = A \cos \theta
A_y = A \sin \theta

Given this, find the components of each force. (I suggest you do some reading on vectors and vector addition. Here's a good place to start: http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#vec2)

 

[ChinToka]

the only angle that is given is 150°, so x and y are "60 cos 150" and "60 sin 150"? I have some trouble identifying all necessary components.

 

[Doc Al]

150° is the angle of the 75 N force, not the 60 N force.

One force is 60N along the x-axis

Thus its x component is 60 N and its y component is 0. (The angle it makes with the x-axis is 0 degrees.)

Now find the components of the 75 N force.

 

[ChinToka]

Allright, the y component is 0 because there is no angle given. Then the y component of 75N is 210 because of the counterclockwise 150°. The only difference i get when i calculate it is with 150° is (-64.95;37.5) and with 210° (-64.95;-37.5). The y component just turned negative.

 

[Doc Al]

Allright, the y component is 0 because there is no angle given.

The y component of the 60 N force is 0 because the angle is 0. (The x-axis is at 0 degrees to itself.)

Then the y component of 75N is 210 because of the counterclockwise 150°.

Huh? How can a component of a vector be greater than the vector itself?

The only difference i get when i calculate it is with 150° is (-64.95;37.5)

These are the correct components of the 75 N force.

and with 210° (-64.95;-37.5).

Where does the 210° come from? The angle is given as 150°.

 

[ChinToka]

my mistake, there were too many numbers flying around in my head

I was able to catch my instructor in his office hours and we solved the problem. BUT he did it too quick for me to understand and the notes he gave me are made by his thoughts, so I can´t follow them. Maybe we can still work on this problem so that I can understand it in full, can we?