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

Commented by mathdave last updated on 16/Feb/24

$$plssomeoneshouldmeout$$

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

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

$$N_1+T\sin \theta =G_1$$$$T\cos \theta =f_{1}with{f}_1=\mu N_1$$$$N_1+G_2=N_2$$$$P=f_1+f_{2}with{f}_2=\mu N_2$$$$$$$$N_1=G_1-T\sin \theta =G_1-f_1\tan \theta =G_1-\mu N_1\tan \theta$$$$\Rightarrow N_1=\frac{G_1}{1+\mu \tan \theta }$$$$N_2=G_2+\frac{G_1}{1+\mu \tan \theta }$$$$P=\mu (G_2+\frac{2G_1}{1+\mu \tan \theta })$$$$=0.3(2+\frac{2\times 1}{1+0.3\times \tan 30°})\approx 1.11KN$$

Commented by mathdave last updated on 16/Feb/24

$$thankssomuchmrW$$

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