AP EAPCET 08-Jul-2022 Shift 1 Paper

Torque on a particle,
τ=r×F
r= position vector of particle
F= Force vector
As particle is located at (1,1,1) m; position vector of particle,
r=x

^

i

+y

^

j

+z

^

k

=(

^

i

+

^

j

+

^

k

)m
And direction of force is along vector (

^

i

+

^

j

)
Hence, unit vector in direction of force,

^

n

=(

^

i

+

^

j

)∕√2
force in vector form,
F= magnitude × direction
=2

^

n

=2

( ^ i + ^ j )

√2

=√2(

^

i

+

^

j

)N
So torque,
τ=r×F
=|

^ i	^ j	^ k
1	1	1
√2	√2	0

|
=

^

i

(1×0−√2×1)−

^

j

(1×0−√2×1) +

^

k

(1×√2−√2×1)
=−√2

^

i

+√2

^

j

+0

^

k

Torque =(−√2

^

i

+√2

^

j

)N−m