Rotational Equilibrium and Newton's First Law in Rotational Form

AP Physics 1· difficulty 3/5

A 2 m2~\text{m} horizontal beam (massless) is supported at both ends. A 300 N300~\text{N} weight is placed 0.5 m0.5~\text{m} from the left support. Force on the <strong>right</strong> support is

  • A

    75 N75~\text{N}

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  • B

    50 N50~\text{N}

  • C

    150 N150~\text{N}

  • D

    225 N225~\text{N}

Explanation

Torques about the left support: FR2=3000.5FR=75 NF_R \cdot 2 = 300 \cdot 0.5 \Rightarrow F_R = 75~\text{N}.

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