Question

A flat, horizontal uniform plank is supported underneath by a triangular wedge which itself rests on flat, level ground. An object that has the same weight as the plank is placed on the left end of the plank. Where the wedge must be placed under the plank in order for the system to be in equilibrium?
A. 1/6 the length of the plank from the left end
B. 1/5 the length of the plank from the left end
C. 1/4 the length of the plank from the left end
D. 1/3 the length of the plank from the left end

Answer 1

Answer:C

Explanation:

Given

weight of object is equal to weight of object

Suppose weight of Planck is W

suppose weight is at distance of x cm from wedge

balancing Torque

$w\times x-w(\frac{L}{2}-x)=0$

$2x=\frac{L}{2}$

$x=\frac{L}{4}$

i.e. at a distance of 0.25L from the Left end  

Answer 2

Answer:

C. 1/4 the length of the plank from the left end

Explanation:

• We consider the beam to be mass-less and having a uniformly distributed load of $(\frac{w}{l} )\ N.m^{-1}$

For a beam in equilibrium we have all the forces and moments on the beam in balanced condition.

Mathematically:

$\sum F=0$

$\sum M=0$

For the forces to be balanced:

$w+(\frac{w}{l} )\times l=R_w$

$R_w=2w$ ..........................(1)

Taking moment about the left end in the balanced condition:

$M_A=0$

$w\times \frac{l}{2} =R_w\times x$

put the value of $R_w$ from eq.(1)

$w\times \frac{l}{2} =2w\times x$

$x=\frac{l}{4}$