MyQuestionIcon

1

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

The Complete Equilibrium

In figure, th...

Question

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

Open in App

Solution

Taking torque about the attachment point for $W$ , we get
$- T_{1} (0.4 L) + T_{2} (0.3 L) + 500 (0.2 L) = 0$
where $T_{1} = T_{2} = T$
$T (0.1) = 100 \Rightarrow T = 1000 N$
By balancing forces
$\sum F_{y} = 0 \Rightarrow 2 T - W - 500 = 0$
$\Rightarrow W = 1500 N$

Suggest Corrections

thumbs-up

59

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 Fig. the bar is uniform and weighing

500 N

. How large must

W

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 ?

Q. In the following figures (i) and (ii), all the surfaces are smooth, string is massless and pulley is ideal. The mass of rope is

m

in both the cases. If

T_{1}

T_{2}

be the tensions at the mid points of the ropes in figure (i) and figure (ii) respectively, then

T_{1} : T_{2}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Equilibrium 2.0

PHYSICS

Watch in App

explore_more_icon

Explore more

The Complete Equilibrium

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon