$3\; m$ long ladder wetghing $20 kg$ leans on a frictionless wall. Its feet rest on the floor $1\; m$ from the wall as shown in Figure Find the reaction forces of the wall and the floor.
$3\; m$ long ladder wetghing $20 kg$ leans on a frictionless wall. Its feet rest on the floor $1\; m$ from the wall as shown in Figure Find the reaction forces of the wall and the floor.
The ladder AB is 3 m long, its foot $A$ is at distance $A C=1$ m from the wall. From Pythagoras theorem, $BC =2 \sqrt{2}$ m. The forces on the ladder are its weight W acting at its centre of gravity D. reaction forces $F_{1}$ and $F_{2}$ of the wall and the floor respectively. Force $F_{1}$, 1 s perpendicular to the wall, since the wail is frictionless. Force $F_{2}$ is resolved into two components, the normal reaction $N$ and the force of friction $F$. Note that $F$ prevents the ladder from sliding away from the wall and is therefore directed toward the wall.
For translational equilibrium, taking the forces in the vertical direction. $N-W=0$
Taking the forces in the horizontal direction.
$F-F_{1}=0$
For rotational equilibrium, taking the moments of the forces about $A$
$2 \sqrt{2} F_{1}-(1 / 2) W=0$
Now $\quad W=20 g =20 \times 9.8 N =196.0 N$
$N=196.0 N$
$F_{1}=W / 4 \sqrt{2}=196.0 / 4 \sqrt{2}=34.6 N$
$F=F_{1}=34.6 N$
$F_{2}=\sqrt{F^{2}+N^{2}}=199.0 N$
The force $F_{2}$ makes an angle $\alpha$ with the horizontal.
$\tan \alpha=N / F=4 \sqrt{2}, \quad \alpha=\tan ^{-1}(4 \sqrt{2}) \approx 80^{\circ}$
Similar Questions
When it is said that body is in mechanical equilibrium ?
When it is said that body is in mechanical equilibrium ?
A small $100$ $g$ sleeve $B$ can slide on a smooth, circular and rigid wire frame $A$ of radius $5$ $m$ placed in vertical place. The wire frame is rotating about its vertical diameter at $2$ $rad/s$. When the sleeve is brought at a particular angular position other than the bottom and the top of the ring, the sleeve will not slide on the wire frame. ......... $N$ is force of interaction between the sleeve and the wire frame at this position.
A small $100$ $g$ sleeve $B$ can slide on a smooth, circular and rigid wire frame $A$ of radius $5$ $m$ placed in vertical place. The wire frame is rotating about its vertical diameter at $2$ $rad/s$. When the sleeve is brought at a particular angular position other than the bottom and the top of the ring, the sleeve will not slide on the wire frame. ......... $N$ is force of interaction between the sleeve and the wire frame at this position.
A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity $\omega$ is an example of a non-intertial fiame of reference. The relationship between the force $\vec{F}_{\text {rot }}$ experienced by a particle of nass in moving on the rotating disc and the force $\vec{F}_{\text {in }}$ experienced by the particle in an inertial frame of reference is
$\vec{F}_{\text {rot }}=\vec{F}_{\text {in }}+2 m\left(\vec{v}_{\text {rot }} \times \vec{\omega}\right)+m(\vec{\omega} \times \vec{r}) \times \vec{\omega},$
where $\vec{v}_{\text {rot }}$ is the velocity of the particle in the rotating frame of reference and $\bar{r}$ is the position vector of the particle with respect to the centre of the disc.
Now consider a smooth slot along a diameter of a disc of radius $R$ rotating counter-clockwise with a constant angular speed $\omega$ about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the $x$-axis along the slot, the $y$-axis perpendicular to the slot and the $z$-axis along the rotation axis $(\vec{\omega}=\omega \hat{k})$. A sm a $1$ block of mass $m$ is gently placed in the slot at $\vec{r}=(R / 2) \hat{i}$ at $t=0$ and is constrained to move only along the slot.
(Image)
($1$) The distance $r$ of the block at time $t$ is
($A$) $\frac{R}{4}\left(e^{\omega t}+e^{-\omega t}\right)$ ($B$) $\frac{R}{2} \cos \omega t$ ($C$) $\frac{R}{4}\left(e^{2 \omega t}+e^{-2 \omega t}\right)$
($D$) $\frac{F}{2} \cos 2 \omega t$
($2$) The net reaction of the disc on the block is
($A$) $\frac{1}{2} m \omega^2 R\left(e^{2 \omega t}-e^{-2 \omega t}\right) \hat{j}+m g \hat{k}$
($B$) $\frac{1}{2} m \omega^2 R\left(e^{\omega t}-e^{-a t t}\right) j+m g k$
($C$) $-m \omega^2 R \cos \omega t \hat{j}-m g \hat{k}$
($D$) $m \omega^2 R \sin \omega t \hat{j}-m g \hat{k}$
Give the answer quetioin ($1$) ($2$)
A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity $\omega$ is an example of a non-intertial fiame of reference. The relationship between the force $\vec{F}_{\text {rot }}$ experienced by a particle of nass in moving on the rotating disc and the force $\vec{F}_{\text {in }}$ experienced by the particle in an inertial frame of reference is
$\vec{F}_{\text {rot }}=\vec{F}_{\text {in }}+2 m\left(\vec{v}_{\text {rot }} \times \vec{\omega}\right)+m(\vec{\omega} \times \vec{r}) \times \vec{\omega},$
where $\vec{v}_{\text {rot }}$ is the velocity of the particle in the rotating frame of reference and $\bar{r}$ is the position vector of the particle with respect to the centre of the disc.
Now consider a smooth slot along a diameter of a disc of radius $R$ rotating counter-clockwise with a constant angular speed $\omega$ about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the $x$-axis along the slot, the $y$-axis perpendicular to the slot and the $z$-axis along the rotation axis $(\vec{\omega}=\omega \hat{k})$. A sm a $1$ block of mass $m$ is gently placed in the slot at $\vec{r}=(R / 2) \hat{i}$ at $t=0$ and is constrained to move only along the slot.
(Image)
($1$) The distance $r$ of the block at time $t$ is
($A$) $\frac{R}{4}\left(e^{\omega t}+e^{-\omega t}\right)$ ($B$) $\frac{R}{2} \cos \omega t$ ($C$) $\frac{R}{4}\left(e^{2 \omega t}+e^{-2 \omega t}\right)$
($D$) $\frac{F}{2} \cos 2 \omega t$
($2$) The net reaction of the disc on the block is
($A$) $\frac{1}{2} m \omega^2 R\left(e^{2 \omega t}-e^{-2 \omega t}\right) \hat{j}+m g \hat{k}$
($B$) $\frac{1}{2} m \omega^2 R\left(e^{\omega t}-e^{-a t t}\right) j+m g k$
($C$) $-m \omega^2 R \cos \omega t \hat{j}-m g \hat{k}$
($D$) $m \omega^2 R \sin \omega t \hat{j}-m g \hat{k}$
Give the answer quetioin ($1$) ($2$)
- [IIT 2016]
Find minimum height of obstacle so that the sphere can stay in equilibrium.
Find minimum height of obstacle so that the sphere can stay in equilibrium.
Two uniform rods of equal length but different masses are rigidly joined to form an $L$ -shaped body, which is then pivoted as shown. If in equilibrium the body is in the shown configuration, ratio $M/m$ will be:
Two uniform rods of equal length but different masses are rigidly joined to form an $L$ -shaped body, which is then pivoted as shown. If in equilibrium the body is in the shown configuration, ratio $M/m$ will be: