A 5.0 kg block at rest on a frictionless surface is acted on by forces. F1+5.5 N and it is a girl pushing on the block with a 30 degree angle. The second force is F2=3.5 N and is a boy pulling on it with a 37 degree angle. What additional force will keep the block at rest?
2007-03-21 10:33:42 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
We are interested in the force that is acting parallel to the ground here. What we have to do, then, is to find out what part of the 5.5N and -3.5N forces are horizontal (3.5 is negative here because it is opposite to the pushing force).
The formula for horizontal forces is the applied force multiplied by the cosine of the angle, or:
5.5*cosine 30
=5.5*.87
=4.79N
and
-3.5*cosine 37
=-3.5*.80
=-2.8
So the girl's horizontal force is 4.79N, and net there are
4.79-2.8=1.99, or about 2 Newtons acting in the horizontal direction (towards the direction the girl is pushing). To counteract this force and have a Net horizontal force of 0 (so the block stays at rest), we need
-1.99Newtons (note the negative sign) towards the opposite direction.
2007-03-21 11:40:45 · answer #1 · answered by bloggerdude2005 5 · 0⤊ 0⤋
First draw free body diagram and resolve the forces
In the x-directon
F1-x=5.5cos(30)=4.8 N
F2-x=-3.5cos(37)=-2.8 N
Total force x-direction=2.0 N
Don't need to worry about Y direction as it is balanced by the surface.
We need to have a force equal to -2 N, or 2 N pulling
2007-03-21 18:39:50 · answer #2 · answered by Rob M 4 · 0⤊ 0⤋
You didn't make it totally clear. The 1st 2 answers assume the boy and girl on the same side of the block fighting over which way it should go. Typical, but maybe not true. The boy could be on the other side helping.
If that's the situation, you can adapt the methods already given.
2007-03-21 19:56:23 · answer #3 · answered by sojsail 7 · 0⤊ 0⤋