Answer:
The torque on the wrench is 4.188 Nm
Explanation:
Let r = xi + yj where is the distance of the applied force to the origin.
Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,
r = 0.18i + 0.055j
The applied force f = 88i - 23j
The torque τ = r × F
So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j
= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j
= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0 since i × i = 0, j × j = 0, i × j = k and j × i = -k
= 0 - 4.14k + 0.0484(-k) + 0
= -4.14k - 0.0484k
= -4.1884k Nm
≅ -4.188k Nm
So, the torque on the wrench is 4.188 Nm
