PartI n an inertial frame of reference, a series of experiments is conducted. In
ID: 1791442 • Letter: P
Question
PartI n an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly move at a constant velocity of 256 m/s? The forces applied are as follows: Check all that apply Hints Hint 1. Using the net force In an inertial frame of reference, moving at a constant velocity is only possible when the net force acting on the object iz zero. 2N: 2 N 200 N: 200 N 200 N: 201 N 2N: 2 N: 4 N 2 N: 2N: 2 N 2N: 2 N: 3 N 2N: 2 N: 5 N 200 N: 200 N: 5NExplanation / Answer
If the object is moving at a constant velocity, then its acceleration is zero. If it is not accelerating, then the net force acting on the object is zero. Solving this problem is therefore equivalent to figuring out which of the cases could be ones in which the net force is zero.
If there are only two forces applied to the object, then for the net force to be zero, the two forces must be equal in magnitude and pointed in opposite directions. The cases (2N, 2N) and (200N, 200N) , where these number represent the magnitude of the applied forces, could possibly be cases in which the net force is zero (if the forces are exactly opposed). The case (200N, 201N) cannot possibly correspond to a case in which the net force is zero.
For the cases with three applied forces, in order for the net force to be zero, it must be possible for the forces to add up to zero when added vectorially. This means that the force "arrows" (where the length of the arrows are the magnitudes of the forces) should be capable of creating a triangle. In geometry, there is a powerful theorem known as the "Triangle Inequality". It states that three line segments of length a, b, and c can only form a triangle if the sum of any two segments is greater than the third segment. (If the sum of two of the segments is *equal* to the third segment, then one has a "degenerate" triangle in which two of the angles are zero, and the third is 180 degrees -- i.e., a line). That is, that a+b >= c, a+c >= b, and b+c >= a. If these inequalities are not satisfied, then the three line segments cannot form a triangle.
The forces (2N, 2N, 4N) can form a degenerate triangle (the two 2N forces acting along the same line, with the 4N force opposing both of them), so it could result in a net force of zero.
(2N, 2N, 2N) obeys the strict (i.e., all inequality signs) triangle inequality, so it could result in a net force of zero. (You can make an equilateral triangle with these sides).
(2N, 2N, 3N) also obeys the strict (i.e., all inequality signs) triangle inequality, so it could result in a net force of zero. (You can make an isosceles triangle with these sides).
(2N, 2N, 5N) does *not* obey the triangle inequality because 2N + 2N is not greater than 5N. It can't result in a net force of zero.
(200N, 200N, 5N) also obeys the strict (i.e., all inequality signs) triangle inequality, so it could result in a net force of zero. (You can make an isosceles triangle with these sides).