One student is pushing on a chair with a force of 260.6 N directed at an angle of 30 degrees above horizontal while a second student pushes on the same side of the chair with a force of 21.0 N at an angle of 15 degrees below horizontal. What is the magnitude, in N, of the sum of these forces?
how do i go about solving this question?
physics - bobpursley, Thursday, October 1, 2009 at 5:39pm
add them.
using i,j coordinates
F= 260.6(sin30 j+ cos30 i)+ 21(sin-15 j+ cos 15 i)
Srry bob but i get a wacky number of number of 236434N when i did it just know.
physics - bobpursley, Thursday, October 1, 2009 at 5:58pm
Find your error.
Or, post your work, and I will check.
physics - tracy, Thursday, October 1, 2009 at 6:05pm
thanks for the quick reply.
Recorrection
F= 260.6(sin30 j+ cos30 i)+ 21(sin-15 j+ cos 15 i)
f= 5,4726( 0.24j + 1.8i)
f=5,4726(0.24)^2 + 1.8 ^2
f=18255.34N
To solve this question, you can use vector addition. Here's how you can go about solving it:
1. Break down each force into its x and y components. To do this, you can use the angles given and the trigonometric identities for sine and cosine. Let's call the first student's force F1 and the second student's force F2.
F1x = F1 * cos(30)
F1y = F1 * sin(30)
F2x = F2 * cos(-15)
F2y = F2 * sin(-15)
2. Calculate the x and y components of the total force by adding the corresponding components of F1 and F2.
Ftotalx = F1x + F2x
Ftotaly = F1y + F2y
3. Calculate the magnitude of the total force using the Pythagorean theorem:
Ftotal = sqrt(Ftotalx^2 + Ftotaly^2)
Plugging in the values, we have:
Ftotalx = 260.6 * cos(30) + 21.0 * cos(-15)
Ftotaly = 260.6 * sin(30) + 21.0 * sin(-15)
Ftotalx = 225.19 N
Ftotaly = -42.36 N
Ftotal = sqrt((225.19)^2 + (-42.36)^2) = 231.4 N
So, the magnitude of the sum of these forces is 231.4 N.