Question-204405

Question Number 204405 by mathdave last updated on 16/Feb/24

Commented by mathdave last updated on 16/Feb/24

p l s s o m e o n e s h o u l d m e o u t

Answered by mr W last updated on 16/Feb/24

Commented by mr W last updated on 16/Feb/24

N_{1} + T \sin θ = G_{1}

T \cos θ = f_{1} w i t h f_{1} = μ N_{1}

N_{1} + G_{2} = N_{2}

P = f_{1} + f_{2} w i t h f_{2} = μ N_{2}

N_{1} = G_{1} - T \sin θ = G_{1} - f_{1} \tan θ = G_{1} - μ N_{1} \tan θ

\Rightarrow N_{1} = \frac{G_{1}}{1 + μ \tan θ}

N_{2} = G_{2} + \frac{G_{1}}{1 + μ \tan θ}

P = μ (G_{2} + \frac{2 G_{1}}{1 + μ \tan θ})

= 0.3 (2 + \frac{2 \times 1}{1 + 0.3 \times \tan 30 °}) \approx 1.11 K N

Commented by mathdave last updated on 16/Feb/24

t h a n k s s o m u c h m r W