Figure shows an overhead view of a 0.026 kg lemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force F1 has a magnitude of 4 N and is at ¥è1 = 27¢ª. Force F2 has a magnitude of 6 N and is at ¥è2 = 25¢ª. If the lemon half is stationary, what are projections of the third force on (a)x-axis, (b)y-axis? If the lemon half has constant velocity v = (12 i - 14 j) m/s, what are projections of the third force on (c)x-axis, (d)y-axis? If the lemon half has varying velocity v = (11ti - 13tj) m/s2, where t is time, what are projections of the third force on (e)x-axis, (f)y-axis?
There is no figure and I do not understand your symbols such as ¥è and ¢ª.
To find the projections of the third force on the x-axis and y-axis, we first need to find the resultant force acting on the lemon half.
To find the resultant force, we can use the vector addition of forces. The resultant force, denoted by R, is given by:
R = F1 + F2 + F3
For the lemon half to be stationary, the resultant force must be zero, implying that the magnitude and direction of the resultant force are both zero. Therefore, we can set up the following equations:
R_x = F1_x + F2_x + F3_x = 0
R_y = F1_y + F2_y + F3_y = 0
a) To find the projection of the third force on the x-axis (F3_x), solve the equation F1_x + F2_x + F3_x = 0 for F3_x.
b) To find the projection of the third force on the y-axis (F3_y), solve the equation F1_y + F2_y + F3_y = 0 for F3_y.
c) Since the lemon half has a constant velocity v = (12i - 14j) m/s, we know that the net force acting on it is zero according to Newton's Second Law (F = ma). Therefore, the vector sum of the forces on the lemon half must be zero:
F1 + F2 + F3 = 0
To find the projections of the third force on the x-axis (F3_x) and y-axis (F3_y), we can set up the following equations:
F1_x + F2_x + F3_x = 0
F1_y + F2_y + F3_y = 0
Solve these equations to find F3_x and F3_y.
d) To find the projections of the third force on the x-axis (F3_x) and y-axis (F3_y) when the lemon half has a varying velocity v = (11ti - 13tj) m/s², we can use Newton's Second Law (F = ma). The net force acting on the lemon half is the derivative of its momentum:
F1 + F2 + F3 = m * (dv/dt)
Differentiate the velocity vector with respect to time to find the acceleration:
a = (dv/dt) = (d/dt)(11ti - 13tj) = 11i - 13j
Substitute the given values of F1, F2, and the acceleration vector into the equation F1 + F2 + F3 = m * a. Then set up the following equations to solve for F3_x and F3_y:
F1_x + F2_x + F3_x = m * (11 i)
F1_y + F2_y + F3_y = m * (-13 j)
Solve these equations to find F3_x and F3_y.
e) The same process as in part d can be used to find the projections of the third force on the x-axis (F3_x) and y-axis (F3_y) when the lemon half has a varying velocity v = (11ti - 13tj) m/s², where t is time.