A horizontal force F is applied at the top of an equilateral triangular block having mass m. The minimum coefficient of friction required to topple the block before translation will be

Question

A horizontal force F is applied at the top of an equilateral triangular block having mass m. The minimum coefficient of friction required to topple the block before translation will be <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-fqpuaglowsajcq7g.jpg" /> (A) (2/√3) (B) (1/2) (C) (1/√3) (D) None of these

Tardigrade Admin · Accepted Answer

I=(m a2/6)+m((a/√3))2⇒ I=((m a2/2)) ⇒ F× (a √3/2)=(m g)(a/2)⇒ F=((m g/√3)) F=Fm a x⇒ (m g/√3)=μ mg⇒ μ =(1/√3) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-fqpuaglowsajhcyv.jpg" />

Q. A horizontal force $F$ is applied at the top of an equilateral triangular block having mass $m .$ The minimum coefficient of friction required to topple the block before translation will be

$\frac{2}{3}$

$\frac{1}{2}$

$\frac{1}{3}$

None of these

$I = \frac{m a ^{2}}{6} + m (\frac{a}{3})^{2} \Rightarrow I = (\frac{m a ^{2}}{2})$
$\Rightarrow F \times \frac{a 3}{2} = (m g) \frac{a}{2} \Rightarrow F = (\frac{m g}{3})$
$F = F_{ma x} \Rightarrow \frac{m g}{3} = μ m g \Rightarrow μ = \frac{1}{3}$

Q. A horizontal force F is applied at the top of an equilateral triangular block having mass m. The minimum coefficient of friction required to topple the block before translation will be

3​2​

21​

3​1​