Resultant Force magnitude of the acceleration.

[BuBbLeS01]

I can't seem to figure this problem out...
Three forces in the x-y plane act on a 3.70 kg mass: 14.50 N directed at 478.0°, 11.00 N directed at 117.0°, and 10.70 N directed at 222.0°. All angles are measured from the positive x-axis, with positive angles in the counter-clockwise direction. Calculate the magnitude of the acceleration.

I am not sure what I am doing wrong here. I am adding the x components and y components of the 3 forces then I am finding the resultant vector. Next I take that and divide it by the mass to get the acceleration, using Newtons 2nd Law...Is that not right?

 

[FedEx]

I can't seem to figure this problem out...
Three forces in the x-y plane act on a 3.70 kg mass: 14.50 N directed at 478.0°, 11.00 N directed at 117.0°, and 10.70 N directed at 222.0°. All angles are measured from the positive x-axis, with positive angles in the counter-clockwise direction. Calculate the magnitude of the acceleration.

I am not sure what I am doing wrong here. I am adding the x components and y components of the 3 forces then I am finding the resultant vector. Next I take that and divide it by the mass to get the acceleration, using Newtons 2nd Law...Is that not right?

Show us your calculations. Without knowing them we cannot help you. It is possible that you might have made some mistake while taking the componenets.

 

[BuBbLeS01]

Show us your calculations. Without knowing them we cannot help you. It is possible that you might have made some mistake while taking the componenets.

I am sorry...lol I guess that would help! I am new! If you couldn't tell hehe..Okay here...
Fx1= 14.5 sin 478
Fx2= 11.0 sin 117
Fx3= 10.7 cos 222
sum= 14.65216223

Fy1= 14.5 cos 478
Fy2= 11.0 cos 117
Fy3= 10.7 sin 222
sum= -18.96093065

14.652^2 + -18.961^2= 574.20
Resultant force= 23.96
a= 23.96 N/3.7 kg

 

[FedEx]

I am sorry...lol I guess that would help! I am new! If you couldn't tell hehe..Okay here...
Fx1= 14.5 sin 478
Fx2= 11.0 sin 117
Fx3= 10.7 cos 222
sum= 14.65216223

Fy1= 14.5 cos 478
Fy2= 11.0 cos 117
Fy3= 10.7 sin 222
sum= -18.96093065

14.652^2 + -18.961^2= 574.20
Resultant force= 23.96
a= 23.96 N/3.7 kg

Its ok. But now coming towards the problem why have you taken sines for the x component. for Fx1 and Fx2 you have taken sine. You have to take cosines. And simplify 478. So we get 478-360=118. Moreover F1 and F2 are in second quadrant and hence cos will be negative and hence you will get Fx1 and Fx2 negative. F3 is in third quadrant so even there Fx3 will be negative.
Turning towards the y components there you have taken cos instead of taking sine. Here as F1 and F2 are in second quadrant and in second quadrant sine is positive so you will be getting positive values for Fy1 and Fy2. Now F3 is in third quadrant and so you will be getting negative Fy3.
Hence you will be getting proper answers.Still if you are having any prob, I am always there.

 

[BuBbLeS01]

Its ok. But now coming towards the problem why have you taken sines for the x component. for Fx1 and Fx2 you have taken sine. You have to take cosines. And simplify 478. So we get 478-360=118. Moreover F1 and F2 are in second quadrant and hence cos will be negative and hence you will get Fx1 and Fx2 negative. F3 is in third quadrant so even there Fx3 will be negative.
Turning towards the y components there you have taken cos instead of taking sine. Here as F1 and F2 are in second quadrant and in second quadrant sine is positive so you will be getting positive values for Fy1 and Fy2. Now F3 is in third quadrant and so you will be getting negative Fy3.
Hence you will be getting proper answers.Still if you are having any prob, I am always there.

ok I am trying to understand this but first of I thought the to find the x components of f1 and f2 we used sin since the angle was opposite of the side?

 

[turdferguson]

For an angle in standard position (between the x-axis and force vector), sine will give the y component and cosine will give the x component. You have to be careful if the question says something like "20 degrees west of north", but here they state each angle is in standard position

 

[FedEx]

ok I am trying to understand this but first of I thought the to find the x components of f1 and f2 we used sin since the angle was opposite of the side?

If you look at the problem graphically than you will notice that the x comp of F1 and F2 are through negative x-axis and that's what we are getting if we apply cos for F1 and F2. As cos 117 and cos118 are negative we will be getting Fx1 and Fx2 negative ie through the negative x axis.
Now try to solve the problem as i have mentioned. You will get your answer and then you will understand it on your own.

 

[BuBbLeS01]

Thank you both so much for your help...I got the answer right :) now I just have to understand the whole sin and cos thing hehe. We have a horrible teacher who is knew at our school and we are all so confused!