Sppose that a point mass 'm' is moving under a constant force `vecF = 2hati-hatj + hatk` netweon . At some instant , t=0, point P(xm, ym, -1m) [m- metre ] is the instantaneous position of the mass. We know that torque can be expressed as the cross- product of position vector and forces vector, i.e.,
`tau= vecr xx vecF` . At P, torque can be expessed as `tau= (-4hatj - 4 hatk)`Nm At some other instant, t=3 sec, the point mass has another instantaneous position `Q(x_(1), y_(1), z_(1))` such that the displacement vectors between points P and Q and the given force are mutually perpendicular. Also, x-component of torqure at Q is zero and y z-components are equal in magnitude and direction along the negative direction of the respective axes. Using a definite scale, if we construct a parallelogram with the position vectors of Q and the gives force `vecF` as its adjacent sides , area of this parallelogram is `2sqrt(2)m^(2)` . Area of the given parallelogram , in fact , represents a physical quantity whose magnitude in SI system can be expressed as 5 times the gives are
Answer the following questions.
At Q torque acting on the mass can be expressed as :
Sppose that a point mass 'm' is moving under a constant force `vecF = 2hati-hatj + hatk` netweon . At some instant , t=0, point P(xm, ym, -1m) [m- metre ] is the instantaneous position of the mass. We know that torque can be expressed as the cross- product of position vector and forces vector, i.e.,
`tau= vecr xx vecF` . At P, torque can be expessed as `tau= (-4hatj - 4 hatk)`Nm At some other instant, t=3 sec, the point mass has another instantaneous position `Q(x_(1), y_(1), z_(1))` such that the displacement vectors between points P and Q and the given force are mutually perpendicular. Also, x-component of torqure at Q is zero and y z-components are equal in magnitude and direction along the negative direction of the respective axes. Using a definite scale, if we construct a parallelogram with the position vectors of Q and the gives force `vecF` as its adjacent sides , area of this parallelogram is `2sqrt(2)m^(2)` . Area of the given parallelogram , in fact , represents a physical quantity whose magnitude in SI system can be expressed as 5 times the gives are
Answer the following questions.
At Q torque acting on the mass can be expressed as :
`tau= vecr xx vecF` . At P, torque can be expessed as `tau= (-4hatj - 4 hatk)`Nm At some other instant, t=3 sec, the point mass has another instantaneous position `Q(x_(1), y_(1), z_(1))` such that the displacement vectors between points P and Q and the given force are mutually perpendicular. Also, x-component of torqure at Q is zero and y z-components are equal in magnitude and direction along the negative direction of the respective axes. Using a definite scale, if we construct a parallelogram with the position vectors of Q and the gives force `vecF` as its adjacent sides , area of this parallelogram is `2sqrt(2)m^(2)` . Area of the given parallelogram , in fact , represents a physical quantity whose magnitude in SI system can be expressed as 5 times the gives are
Answer the following questions.
At Q torque acting on the mass can be expressed as :
Text Solution
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The correct Answer is:
C
`" "tau_(p) = vecr_(po) xx vecF`
`" "= ("xi" + "yj"-hatk)xx (2i-j+hatk)`
`" "vectau_(p)=(y -1)i -(x + 2)j-(x +2y)hatk`
Given `" "vectau_(p) = - 4hatj - 4hatk`
`therefore" "x=2 and y=1`
Also, `" "vecr_(PQ)=(x_(1) -x) i + (y_(1) - y)j +(z_(1) + 1)hatk`
Given `" "vecr_(PQ) botvecF`
`therefore" "vecr_(PQ) .vecF = 0`
`therefore[(x_(1) - 2)hati + (y_(1) - 1) hatj + ( z_(1)+ 1)hatk].[2i - hatj+hatk]= 0......(i)`
`" "tau_(Q)= vecr_(QO) xxvecF= |{:("i","j","k"),(x_(1),y_(1),z_(1)),(2,-1,1):}|`
`tau_(Q)= (y_(1) + z_(1))hati - j (x_(1) -2z_(1))+ k(-x_(1)- 2y_(1))`
Given, `" "y_(1) + z_(1) =0" "...........(ii)`
And, `" "x_(1) -2z_(1) =x_(1) + 2y_(1)" "......(iii)`
Also, `" "|vecr_(QO) xx vecF= 10sqrt(2)`
`therefore (y_(1) + z_(1))^(2) + (x_(1) -2z_(1))^(2) + (x_(1) + 2y_(1))^(2) = 100 xx 2" "....(iv)`
`therefore " "2(x_(1) - 2z_(1))^(2) = 100 xx 2`
`therefore" "x_(1) - 2z_(1) = 10`
`therefore" "vectau_(Q) = - 10 hatj - 10 hatk`
`" "W = 0, " as " vecr_(PQ)` is prependicular to `vecF`
`" "= ("xi" + "yj"-hatk)xx (2i-j+hatk)`
`" "vectau_(p)=(y -1)i -(x + 2)j-(x +2y)hatk`
Given `" "vectau_(p) = - 4hatj - 4hatk`
`therefore" "x=2 and y=1`
Also, `" "vecr_(PQ)=(x_(1) -x) i + (y_(1) - y)j +(z_(1) + 1)hatk`
Given `" "vecr_(PQ) botvecF`
`therefore" "vecr_(PQ) .vecF = 0`
`therefore[(x_(1) - 2)hati + (y_(1) - 1) hatj + ( z_(1)+ 1)hatk].[2i - hatj+hatk]= 0......(i)`
`" "tau_(Q)= vecr_(QO) xxvecF= |{:("i","j","k"),(x_(1),y_(1),z_(1)),(2,-1,1):}|`
`tau_(Q)= (y_(1) + z_(1))hati - j (x_(1) -2z_(1))+ k(-x_(1)- 2y_(1))`
Given, `" "y_(1) + z_(1) =0" "...........(ii)`
And, `" "x_(1) -2z_(1) =x_(1) + 2y_(1)" "......(iii)`
Also, `" "|vecr_(QO) xx vecF= 10sqrt(2)`
`therefore (y_(1) + z_(1))^(2) + (x_(1) -2z_(1))^(2) + (x_(1) + 2y_(1))^(2) = 100 xx 2" "....(iv)`
`therefore " "2(x_(1) - 2z_(1))^(2) = 100 xx 2`
`therefore" "x_(1) - 2z_(1) = 10`
`therefore" "vectau_(Q) = - 10 hatj - 10 hatk`
`" "W = 0, " as " vecr_(PQ)` is prependicular to `vecF`
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Sppose that a point mass 'm' is moving under a constant force vecF = 2hati-hatj + hatk netweon . At some instant , t=0, point P(xm, ym, -1m) [m- metre ] is the instantaneous position of the mass. We know that torque can be expressed as the cross- product of position vector and forces vector, i.e., tau= vecr xx vecF . At P, torque can be expessed as tau= (-4hatj - 4 hatk) Nm At some other instant, t=3 sec, the point mass has another instantaneous position Q(x_(1), y_(1), z_(1)) such that the displacement vectors between points P and Q and the given force are mutually perpendicular. Also, x-component of torqure at Q is zero and y z-components are equal in magnitude and direction along the negative direction of the respective axes. Using a definite scale, if we construct a parallelogram with the position vectors of Q and the gives force vecF as its adjacent sides , area of this parallelogram is 2sqrt(2)m^(2) . Area of the given parallelogram , in fact , represents a physical quantity whose magnitude in SI system can be expressed as 5 times the gives are Answer the following questions. Work done the for the motion of the points mass from P to Q is :
Sppose that a point mass 'm' is moving under a constant force vecF = 2hati-hatj + hatk netweon . At some instant , t=0, point P(xm, ym, -1m) [m- metre ] is the instantaneous position of the mass. We know that torque can be expressed as the cross- product of position vector and forces vector, i.e., tau= vecr xx vecF . At P, torque can be expessed as tau= (-4hatj - 4 hatk) Nm At some other instant, t=3 sec, the point mass has another instantaneous position Q(x_(1), y_(1), z_(1)) such that the displacement vectors between points P and Q and the given force are mutually perpendicular. Also, x-component of torqure at Q is zero and y z-components are equal in magnitude and direction along the negative direction of the respective axes. Using a definite scale, if we construct a parallelogram with the position vectors of Q and the gives force vecF as its adjacent sides , area of this parallelogram is 2sqrt(2)m^(2) . Area of the given parallelogram , in fact , represents a physical quantity whose magnitude in SI system can be expressed as 5 times the gives are Answer the following questions. Coordinates of P are :
Knowledge Check
Find the torque of a force vecF=2hati+hatj+4hatk acting at the point vecr=7hati+3hatj+hatk :
Find the torque of a force vecF=2hati+hatj+4hatk acting at the point vecr=7hati+3hatj+hatk :
A
`14hati-38hatj-16hatk`
B
`4hati+4hatj+6hatk`
C
`-14hati+38hatj-16hatk`
D
`11hati-26hatj+hatk`
The torque of force F =(2hati-3hatj+4hatk) newton acting at the point r=(3hati+2hatj+3hatk) metre about origin is (in N-m)
The torque of force F =(2hati-3hatj+4hatk) newton acting at the point r=(3hati+2hatj+3hatk) metre about origin is (in N-m)
A
`6hati-6 hatj+12 hatk`
B
`17 hati-6hatj-13hatk`
C
`-6hati+6hatj-12hatk`
D
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