Answer :
Answer:
τ = 0
Explanation:
At the moment it is defined
τ = F x r
In tete case they give us the strength and position in Cartesian form, so it is easier to solve the determinant
τ = [tex]\left[\begin{array}{ccc}i&j&k\\F_{x}&F_{y}&F_{z}\\x&y&z\end{array}\right][/tex]
Let's apply this expression to the exercise
a) P = (-6 i ^ -3j ^ -6 k ^) m
F = (-6 i ^ -3j ^ -6k ^) 103 N
τ =[tex]\left[\begin{array}{ccc}i&j&k\\-6&-3&-6\\-6&-3&-6\end{array}\right][/tex]
τ = i ^ (3 6 - 3 6) + j ^ (6 6 -6 6) + k ^ (6 3 - 3 6)
τ = 0
b) P = 24i ^ + 8j ^ + 9k ^
F = 24i + 8j + 9k
τ = i ^ (72-72) + j ^ (216-216) + k ^ (24 8 - 8 24)
τ = 0
c) P = -6i + 6j-4k
F = -6i + 6j-4k
τ = i ^ (24-24) + j ^ (- 24 + 24) + k ^ (-36 + 36)
τ = 0
.d) P = 24i-8j + 9k
Let's change the sign of strength
F = -24i + 8j-9k
Tae = (I j k 24 -8 9 -24 8 -9)
Tae = i ^ (72 -72) + j ^ (- 216 + 216) + k ^ (192-192)
Tae = 0