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Question

A block of mass M=50 kg, base b=1 m and height h=3 m is kept on a rough inclined surface with coefficient of static friction μ=0.8 as shown in figure. The angle of inclination with horizontal is 37. Determine whether the block slides down or topples over.



A
slides down
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B
topples over
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C
remains in complete equilibrium
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D
cannot be determined.
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Solution

The correct option is B topples over
Coefficient of static friction, μ=0.8
Here, bh=13=0.33
where b= base and h= height of block.
tanθ=tan37=34=0.75
Condition for no slipping is,
μtanθ ...(i)
Condition for toppling is,
μbh ...(ii)

From the Eq. (i) and (ii), we find that as per given data, μ>tanθ & μ>bh, hence block will have no sliding tendency, rather it will tend to topple over.

Method II:
The normal reaction N will shift towards left to counter the torque of friction, i.e at distance x from centre.
For translational equilibrium:
N=Mgcos37
N=50×10×45=400 N
f=Mgsin37
f=50×10×35=300 N
fmax=μN=0.8×400=320 N

Clearly ffmax,supports no sliding of block.

For rotational equilibrium, net torque about centre of block should be zero.
τC=0 ...(iii)
τ=F×r, considering anticlockwise sense of rotation for torque as +ve
τN=(N×x), τf=+f(h2)
& τMg=0 as Mg passes through centre C.

Substituting in Eq (iii),
τMg+τN+τf=0
Nx+(f×h2)=0
x=fh2N=300×32×400
x=98 m
The value of x cannot be more than the value of the base of the block b=1 m
value of x, justifies that block will topple over.

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