Find your error.
Or, post your work, and I will check.
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.
3 answers
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
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
well, the first step to the second is wrong, there is no way 260+21N became 54726 Newtons.
I don't follow your last step (didn't punch it in the calc), but you have to take sqrt of the sum of forces, I don't see where you did that. Resultant= sqrt(xforeces + y forces)
I don't follow your last step (didn't punch it in the calc), but you have to take sqrt of the sum of forces, I don't see where you did that. Resultant= sqrt(xforeces + y forces)
1 answer