There are two forces on the 1.18 kg box in the overhead view of the figure but o
ID: 1878778 • Letter: T
Question
There are two forces on the 1.18 kg box in the overhead view of the figure but only one is shown. For F1 = 19.3 N, a = 10.7 m/s2, and = 24.0°, find the second force (a) in unit-vector notation and as (b) a magnitude and (c)a direction. (State the direction as a negative angle measured from the +x direction.) There are two forces on the 1.18 kg box in the overhead view of the figure but only one is shown. For F1 = 19.3 N, a = 10.7 m/s2, and = 24.0°, find the second force (a) in unit-vector notation and as (b) a magnitude and (c)a direction. (State the direction as a negative angle measured from the +x direction.)Explanation / Answer
Let us assume the second force as F2.
Now write the famous expression of force and acceleration -
F = ma
We expand this to include an unknown force F2, which is added to the force F1 -
F1 + F2 = ma
=> F2 = ma - F1
=> F2 = 1.18 * a - 19.3
We then just plug in the 'a' value and find the resultant Fx value.
(a) Given that, a = 10.7 m/s^2
So, F2 = 1.18 * 10.7 - 19.3 = 12.63 - 19.30 = -6.67 N
The result is negative so, F2 acts in the opposite direction as the assumed positive force F1.
So, the vector in unit vector notation is -
F2 = [6.67*cos(90+24) i + 6.67*sin(90+24) j ] N
= [6.67*cos(114) i + 6.67*sin(114) j ] N = (-2.71i + 6.09j) N
(b) As we have calculated above, magnitude of F2 -
|F2| = 6.67 N
(c) Direction in negative angle from + x direction = 114 - 360 = -246.0 deg.