A uniform metal plate shaped like a triangle | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The triangle $ABC$ has sides $AB=3 \,cm$,$AC=4 \,cm$,and $BC=5 \,cm$. Since $3^2 + 4^2 = 5^2$,it is a right-angled triangle with the right angle a

Vedclass · Accepted Answer

$140 \,g$ at $B$

Vedclass · Answer

(C) The triangle $ABC$ has sides $AB=3 \,cm$,$AC=4 \,cm$,and $BC=5 \,cm$. Since $3^2 + 4^2 = 5^2$,it is a right-angled triangle with the right angle at $A$.Let the mass of the plate be $M = 540 \,g$. The weight of the plate acts through its centroid $G$.To keep the side $BC$ horizontal,the torque about the pivot $A$ due to the weight of the plate must be balanced by the torque due to an additional mass $m_1$ placed at a vertex.Let $AE$ be the altitude from $A$ to $BC$. In $	riangle ABC$,$AE = (AB 	imes AC) / BC = (3 	imes 4) / 5 = 2.4 \,cm$.$BE = \sqrt{AB^2 - AE^2} = \sqrt{3^2 - 2.4^2} = \sqrt{9 - 5.76} = \sqrt{3.24} = 1.8 \,cm$.$EC = BC - BE = 5 - 1.8 = 3.2 \,cm$.The horizontal distance of the centroid $G$ from the vertical line passing through $A$ is the horizontal distance of $G$ from $AE$,which is $GH$. Since $G$ is the centroid,its distance from $BC$ is $1/3$ of the altitude $AE$,and its horizontal distance from $AE$ is $1/3$ of the distance of the centroid of the base $BC$ from $E$. The centroid $G$ divides the median $AD$ in ratio $2:1$. The horizontal distance $GH = (1/3) 	imes (EC - BE) / 2 = (1/3) 	imes (3.2 - 1.8) / 2 = 1.4 / 6 = 7/30 \,cm$.For equilibrium,$M 	imes g 	imes GH = m_1 	imes g 	imes BE$.$540 	imes (7/30) = m_1 	imes 1.8$.$18 	imes 7 = m_1 	imes 1.8$.$126 = 1.8 	imes m_1 \Rightarrow m_1 = 126 / 1.8 = 70 \,g$. Wait,re-evaluating the geometry: The horizontal distance of $G$ from $A$ is the projection of the vector $\vec{AG}$ onto the horizontal. The horizontal distance of $G$ from the vertical line through $A$ is $GH = (1/3) 	imes (BE + EC)/2$ is incorrect. The horizontal distance of $G$ from $A$ is $1/3$ of the horizontal distance of $B$ and $C$ from $A$. The horizontal distance of $B$ from $A$ is $BE = 1.8 \,cm$ and $C$ from $A$ is $EC = 3.2 \,cm$. The centroid $G$ is at a horizontal distance $(1.8 - 3.2)/3 = -1.4/3$ from $A$. To balance,we place $m_1$ at $B$. Torque balance: $M 	imes (1.4/3) = m_1 	imes 1.8 \Rightarrow 540 	imes 1.4 / 3 = m_1 	imes 1.8 \Rightarrow 180 	imes 1.4 = m_1 	imes 1.8 \Rightarrow 252 = 1.8 	imes m_1 \Rightarrow m_1 = 140 \,g$ at $B$.

Similar Questions

$A$ uniform rod of mass $15\,kg$ and length $5\,m$ is held stationary with the help of a light string as shown. The tension in the string is ......... $N.$

$A$ metre stick is balanced on a knife edge at its centre. When two coins,each of mass $5\; g$ are put one on top of the other at the $12.0 \;cm$ mark,the stick is found to be balanced at $45.0\; cm$. What is the mass of the metre stick?

$A$ heavy iron bar of mass $12 \,kg$ has one end on the ground and the other on the shoulder of a man. The rod makes an angle $60^{\circ}$ with the horizontal. The weight experienced by the man is:

In an experiment with a beam balance,an unknown mass $m$ is balanced by two known masses of $16\, kg$ and $4\, kg$ as shown in the figure. The value of the unknown mass $m$ is ....... $kg$.

In a physical balance working on the principle of moments,when $5\, mg$ weight is placed on the left pan,the beam becomes horizontal. Both the empty pans of the balance are of equal mass. Which of the following statements is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module