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The minimum force required to slide a body of weight 'W' on a rough horizontal plane is

A. W sinθ

B. W cosθ

C. W tanθ

D. None of these

Answer: Option C


This Question Belongs to Mechanical Engineering >> Engineering Mechanics

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Comments ( 8 )

  1. KriShh
    KriShh :
    2 years ago

    You are misleading by Copying the answer from Khurmi text book. Here correct answer is option A. Please correct it.

  2. Mohammad Nasir
    Mohammad Nasir :
    3 years ago

    A is correct , plz make it correct

  3. TARIQ AHMAD
    TARIQ AHMAD :
    3 years ago

    A is correct

  4. Chandu Smiley
    Chandu Smiley :
    3 years ago

    A is correct.. As force acting at an angle R=w-psintetha
    Then p= wsinthetha is the min. Force required

  5. Nikunj Miyani
    Nikunj Miyani :
    3 years ago

    Ans A is correct

  6. Ashish Mishra
    Ashish Mishra :
    3 years ago

    Answer is Wsinθ. Please correct.

  7. Sarveshwer Chandra
    Sarveshwer Chandra :
    4 years ago

    A. W sinθ

  8. Madhavan Palmani
    Madhavan Palmani :
    4 years ago

    P = Wsinfi/cos(θ-fi).
    We know that, angle of inclinstion = angle of friction for rough horizontal plane. So that answer is Wsin(θ)

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If a number of forces are acting at a point, their resultant is given by

A. $${\left( {\sum {\text{V}} } \right)^2} + {\left( {\sum {\text{H}} } \right)^2}$$

B. $$\sqrt {{{\left( {\sum {\text{V}} } \right)}^2} + {{\left( {\sum {\text{H}} } \right)}^2}} $$

C. $${\left( {\sum {\text{V}} } \right)^2} + {\left( {\sum {\text{H}} } \right)^2} + 2\left( {\sum {\text{V}} } \right)\left( {\sum {\text{H}} } \right)$$

D. $$\sqrt {{{\left( {\sum {\text{V}} } \right)}^2} + {{\left( {\sum {\text{H}} } \right)}^2} + 2\left( {\sum {\text{V}} } \right)\left( {\sum {\text{H}} } \right)} $$