A solid cone hangs from a frictionless pivot at the origin O as shown above. If î, ĵ and k̂ are unit vectors, and a, b, and c are positive constants, which of the following forces F applied to the rim of the cone at point P results in a torque τ on the cone with a negative component τz?
A. F = ak̂, P is (0, b, −c)
B. F = −ak̂, P is (0, −b, −c)
C. F = aĵ, P is (–b, 0, −c)
D. F = aĵ, P is (b, 0, −c)
E. F = −ak̂, P is (−b, 0, −c)
(GR9277 #08)
Solution:
Torque:
Result desired: negative component of τz so we are looking for minus k̂ component.
For k̂ component: rxFy − ryFx
Thus, Fx and Fy cannot be zero
(A) FALSE
F = ak̂ → Fx = 0 and Fy = 0
(B) and (E) FALSE
F = −ak̂ → Fx = 0 and Fy = 0
(C) TRUE
F = aĵ → Fx = 0, Fy = a, rx = −b, ry = 0,
P is (–b, 0, −c) → rx = −b, ry = 0
rxFy − ryFx = −ab − 0 = −abk̂ (negative k̂ component)
(D) FALSE
F = aĵ → Fx = 0, Fy = a,
P is (b, 0, −c) → rx = b, ry = 0,
→ rxFy − ryFx = ab − 0 = abk̂ (positive k̂ component)
Answer: C
For k̂ component: rxFy − ryFx
(A) FALSE
F = ak̂ → Fx = 0 and Fy = 0
(B) and (E) FALSE
F = −ak̂ → Fx = 0 and Fy = 0
(C) TRUE
F = aĵ → Fx = 0, Fy = a, rx = −b, ry = 0,
P is (–b, 0, −c) → rx = −b, ry = 0
rxFy − ryFx = −ab − 0 = −abk̂ (negative k̂ component)
(D) FALSE
F = aĵ → Fx = 0, Fy = a,
P is (b, 0, −c) → rx = b, ry = 0,
→ rxFy − ryFx = ab − 0 = abk̂ (positive k̂ component)
Answer: C
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