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Thu 29 Sep, 2005 12:39 pm
A 5.0 kg block at rest on a frictionless surface is acted on by forces F1=5.5 N at 30 degrees above the -x-axis and F2=3.5 N at 37 degrees above the +x-axis. What additional force will keep the block at rest?
You have to break up each force into its horizontal and vertical components. If F1 is at 30 degrees above the -x axis and I understand what that means, there is a 5.5*sin(30) = 2.75N force downward and a 5.5*cos(30) = 4.76N force in the positive x direction. For F2, the downward force is 3.5*sin(37) = 2.11N and the force in the negative x direction is 2.80N. The surface the block is sitting on is resisting all the vertical forces, so you can ignore them. The imbalance comes from the 4.76N force in the +x direction overcoming the 2.80N force in the -x direction. You need another 1.96N force in the -x direction.