A particle is located at the vector position =(8.00 + 9.00) m and a force exerte

Question

A particle is located at the vector position =(8.00 + 9.00) m and a force exerted on it is given by =(6.00 + 5.00) N.
(a) What is the torque acting on the particle about the origin?
( N · m)
(b) Consider another point about which the torque caused by this force on this particle will be in the opposite direction and half as large in magnitude. Select the following conditions that are true.
No such point can exist.
Only one such point can exist.
Multiple such points can exist.
No such a point can lie on the y-axis.
Only one such point can lie on the y-axis.
Multiple such points can lie on the y-axis.


Determine the position vector of such a point.
m

Explanation / Answer

(a)torque is T=Fxr
so T=(6i+5j)x(8i+9j)
=(54-40)k
=14k
(b)for the torque to be half as before and in opposite direction let us choose a point (x,y)
so (6i+5j)x(xi+yj)=-7k
=>6y-5j=-7
this is the equation of a straight line so there are MULTIPLE POINTS CAN EXIST,ONLY ONE SUCH POINT CAN LIE ON THE Y-AXIS

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