Magnitude and Direction of Puck on Ice

[cupra]

Homework Statement

A 0.170 kg puck rests on frictionless ice. Two hockey players strike the puck simultaneously. #1 force is 350. N exerted at 020.0T, and the other #2 force is 600. N exerted at 065.0T. What is the magnitude and direction of the puck's acceleration at the instant that it is struck.

Homework Equations

The Attempt at a Solution

I have resolved the x and y components of the #1 force using trigonometry (x=119.7N and y=328.9N). I also resolved the x and y components of the #2 force using trigonometry (x=543.8N and y=253.6N). What do I do next?

 

[cupra]

Please look at this I think I'm Close

 

[LowlyPion]

Homework Statement

A 0.170 kg puck rests on frictionless ice. Two hockey players strike the puck simultaneously. #1 force is 350. N exerted at 020.0T, and the other #2 force is 600. N exerted at 065.0T. What is the magnitude and direction of the puck's acceleration at the instant that it is struck.

The Attempt at a Solution

I have resolved the x and y components of the #1 force using trigonometry (x=119.7N and y=328.9N). I also resolved the x and y components of the #2 force using trigonometry (x=543.8N and y=253.6N). What do I do next?

I am going to assume that 20.0T here means the angle in degrees to the X-axis which looks like the way you calculated it anyway.

As you have figured out the forces are vectors. What you are trying to figure is the resulting vector. Hence you have separated them into their components ... so just add them together. X's to x's and y's to y's. Then figure the magnitude of the force from the Root of the sum of the squares. Divide by mass to get a.

Your angle will be given by the ratio of the sides.

 

[cupra]

So, Sum of x1 and x2 = 663.5 and Sum of y1 and y2 = 582.5 therefore the hyp. is 882.9
F=ma, a= 5,193.5 m/s2 for my acceleration. This seems bigger than I thought it would be.
How do I figure the direction? I just don't see it in my mind... And THANKS SO MUCH

 

[LowlyPion]

So, Sum of x1 and x2 = 663.5 and Sum of y1 and y2 = 582.5 therefore the hyp. is 882.9
F=ma, a= 5,193.5 m/s2 for my acceleration. This seems bigger than I thought it would be.
How do I figure the direction? I just don't see it in my mind... And THANKS SO MUCH

What is the ratio of y divided by x? Isn't that Tanθ so if you have the value of Tanθ then to find θ just take the ArcTan in degrees of Y/X . That yields θ doesn't it?