three vector f1=2i-3j -4k, f2=3i+4j-0k,f3=3i-j+5k act on a body of mass 40g causing it to displace for 10sec from p1 ( 2,3,7 ) to p2 ( 3,6,7 ).
(i) estimate the over all displacement vector.
(ii) estimate the resultant force on the body .
(iii) what is the distance from p1 to p2.
(iv) calculate the work done on the body.
(v)in which plane is the body moving .
To solve this problem, we will use vector operations and formulas. Let's go through each part step by step:
(i) To find the overall displacement vector, we subtract the initial position vector from the final position vector. Given that p1 is (2, 3, 7) and p2 is (3, 6, 7), we can find the displacement vector Δp by subtracting p1 from p2:
Δp = p2 - p1 = (3, 6, 7) - (2, 3, 7) = (1, 3, 0)
So, the overall displacement vector is (1, 3, 0).
(ii) To estimate the resultant force on the body, we need to sum up the individual forces acting on it. Given f1, f2, and f3, we can find the resultant force vector, F, by adding them together:
F = f1 + f2 + f3 = (2i - 3j - 4k) + (3i + 4j) + (3i - j + 5k) = (8i + 8j + i) = 9i + 8j - 3k
So, the estimated resultant force vector on the body is 9i + 8j - 3k.
(iii) The distance between two points can be calculated using the distance formula. Given p1 and p2, the distance from p1 to p2 is given by the magnitude of the displacement vector:
Distance = |Δp| = sqrt((1)^2 + (3)^2 + (0)^2) = sqrt(1 + 9 + 0) = sqrt(10)
So, the distance from p1 to p2 is sqrt(10).
(iv) The work done on an object can be calculated using the formula W = F · d, where F is the force vector and d is the displacement vector. In this case, the work done is:
Work = F · Δp = (9i + 8j - 3k) · (1i + 3j + 0k) = (9)(1) + (8)(3) + (-3)(0) = 9 + 24 + 0 = 33
So, the work done on the body is 33.
(v) To determine the plane in which the body is moving, we need to find the cross product of the two displacement vectors. The cross product of two vectors gives a vector that is orthogonal (perpendicular) to both of them.
Let's assume the displacement between p1 and p2 is vector A = (1, 3, 0). Since the body is moving along this displacement vector, it will be in the plane defined by the cross product of A with any of the given force vectors (f1, f2, f3). Let's choose f1.
The cross product of A and f1 will give us a vector perpendicular to both A and f1, which represents the plane in which the body is moving.
A × f1 = (1, 3, 0) × (2, -3, -4)
= (3)(-4) - (0)(-3), (0)(2) - (1)(-4), (1)(-3) - (3)(2)
= (-12, 0, -9)
So, the body is moving in a plane described by the vector (-12, 0, - 9).