CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Torque about a Point

In Fig. the b...

Question

In Fig. the bar is uniform and weighing $500 N$ . How large must $W$ be in $N$ if $T_{1}$ and $T_{2}$ are to be equal ?

Open in App

Solution

Taking torque about the attachment point for $W$ , we get
$- T_{1} (0.4 L) + T_{2} (0.3 L) + 500 (0.2 L) = 0$
$T = 1000 N$ , where $T_{1} = T_{2} = T$
$\sum F_{y} = 0 \Rightarrow 2 T - W - 500 = 0$
$\Rightarrow W = 1500 N$

Suggest Corrections

thumbs-up

0

Similar questions

Q. In figure, the bar is uniform and weighing

500 N

. How large must

W in N

T_{1}

T_{2}

are to be equal?

Q. In figure, the bar is uniform and weighing

500 N

. How large must

W in N

T_{1}

T_{2}

are to be equal?

Q. In the fig the bar is uniform and weighting

500 N

. How large must W be if

T_{1}

T_{2}

Q. In Fig. the bar is uniform and weighing

500 N

. How large must

W

N

T_{1}

T_{2}

are to be equal ?

T_{1}, T_{2}, T_{3}, \dots

be terms of an A.P. If

S_{1} = T_{1} + T_{2} + T_{3} + \dots + T_{n}

S_{2} = T_{2} + T_{4} + T_{6} + \dots + T_{n - 1}

n

is odd, then the value of

\frac{S_{1}}{S_{2}}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

PHYSICS

Watch in App

explore_more_icon

Explore more

Torque about a Point

Standard XI Physics

Join BYJU'S Learning Program

CrossIcon