DETERMINE THE MAGNITUDE OF THE MOMENT OF THE FORCE FRIGHTARR

DETERMINE THE MAGNITUDE OF THE MOMENT OF THE FORCE F^RIGHTARROW = {300 i^cap - 200 j^cap + 150 k^cap}N ABOUT THE X-AXIS. (FOLLOW THE PLAIN). r^rightarrow_oc = {-----------------------------------}-r_0 U^cap _x = ------------------ (APPLY THE SCALAR TRIPLE PRODUCT) M_x = ------------------ N.m

Solution

Given that force F = (300 i - 200 j + 150 k) N

The distance vector of the force F from X axis is

r_F = (0.4 j - 0.2 k) m

Moment of the force F about the X axis is

M_F = r_Fx F = {0.4 j - 0.2 k} x {300 i - 200 j + 150 k} Nm

===> M_F = {(300*0.4) j x i + (150*0.4) j x k - (300*0.2) k x i - (200*0.2) k x j} Nm

===> M_F = M_x = {100 i - 60 j -120 k} Nm

Magnitude of the moment of the force about X axis is

M_F = M_x = {100^² + (-60)² + (-120)²} Nm

===> M_F = M_x = 167.332 Nm