Pressure

AP Physics 1· difficulty 3/5

A₁ (small) A₂ (large)

A hydraulic system has piston areas A1A_1 (small) and A2=4A1A_2 = 4 A_1 (large). To lift a load of 400 N400~\text{N} on the large piston, the force needed on the small piston is

  • A

    400 N400~\text{N}

  • B

    100 N100~\text{N}

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  • C

    25 N25~\text{N}

  • D

    1600 N1600~\text{N}

Explanation

Pascal's principle: pressure equal. F1/A1=F2/A2F1=F2(A1/A2)=400/4=100 NF_1/A_1 = F_2/A_2 \Rightarrow F_1 = F_2 (A_1/A_2) = 400/4 = 100~\text{N}.

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