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Two forces 50N and 100 N are acting on a...

Two forces `50N and 100 N` are acting on a rod capable of rotating it about `O` as shown in Fig. What is the net torque acting on the rod ?

Text Solution

Verified by Experts

In Fig. we have resolved force of `100 N` into two rectangular components : `100 cos 30^(@)` along the horizontal and `100 sin 30^(@)` along the vertical.
The horizontal component cannot rotate the rod.
Torque due to `F = 100 sin 30^(@)` about fixed `O`
`tau_(1) = 100 sin 30^(@) xx 0.5 = 100 xx (1)/(2) xx 0.5 = 25 Nm`
It is in clockwise direction.
Torque due to `F = 50N` about fixed point `O`
`tau_(2) = F xx 0.5 = 50 xx 0.5 = 25 Nm`
It is also clockwise.
`:.` Net torque acting on the rod `= tau_(1) + tau_(2) = 25 + 25 = 50Nm`, clockwise.
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Knowledge Check

  • Two forces F_(1) " and " F_(2) are acting on a rod abc as shown in figure.

    A
    if `F_(1)=F_(2)` then `tau_(a)=tau_(b)=tau_(c)` ( for both forces)
    B
    if `F_(1)=F_(2)` then `tau_(a)=tau_(c)netau_(b)` ( for both forces)
    C
    if `F_(1)neF_(2)` then `tau_(a)netau_(b)netau_(c)` ( for both forces)
    D
    if `F_(1)neF_(2)` then `tau_(a)=tau_(c)netau_(b)` ( for both forces)

  • A torque of 100 N-m acting on a wheel at rest, rotates it through 200 radian in 10 s. What is the moment of inertia of the wheel?

    A
    `"10 kg-m"^(2)`
    B
    `"15 kg-m"^(2)`
    C
    `"20 kg-m"^(2)`
    D
    `"25 kg-m"^(2)`

  • If a force 200N acts on a body, the change in momentum is 100 kg m/s. What is the time for which the force acts on th body?

    A
    1 sec
    B
    0.75 sec
    C
    0.5 sec
    D
    0.25 sec