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A body of mass 10 kg is lying on a rough...

A body of mass 10 kg is lying on a rough plane inclined at an angle of `30^(@)` to the horizontal and the coefficient of friction is 0.5. the minimum force required to pull the body up the plane is

A

914 N

B

91.4 N

C

9.14 N

D

0.914 N

Text Solution

AI Generated Solution

The correct Answer is:
To find the minimum force required to pull a body of mass 10 kg up a rough inclined plane at an angle of 30 degrees with a coefficient of friction of 0.5, we can follow these steps: ### Step 1: Identify the forces acting on the body 1. **Weight (W)**: The weight of the body can be calculated using the formula: \[ W = mg \] where \( m = 10 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). ### Step 2: Calculate the weight of the body \[ W = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 98 \, \text{N} \] ### Step 3: Resolve the weight into components 1. **Parallel to the incline (W_parallel)**: This component acts down the incline and is given by: \[ W_{\text{parallel}} = W \sin \theta = mg \sin \theta \] 2. **Perpendicular to the incline (W_perpendicular)**: This component acts normal to the incline and is given by: \[ W_{\text{perpendicular}} = W \cos \theta = mg \cos \theta \] ### Step 4: Calculate the components of the weight 1. For \( \theta = 30^\circ \): \[ W_{\text{parallel}} = 98 \, \text{N} \times \sin(30^\circ) = 98 \, \text{N} \times 0.5 = 49 \, \text{N} \] \[ W_{\text{perpendicular}} = 98 \, \text{N} \times \cos(30^\circ) = 98 \, \text{N} \times \frac{\sqrt{3}}{2} \approx 84.87 \, \text{N} \] ### Step 5: Calculate the frictional force (F_friction) 1. The frictional force opposing the motion can be calculated using: \[ F_{\text{friction}} = \mu \cdot N \] where \( N = W_{\text{perpendicular}} \) and \( \mu = 0.5 \). \[ F_{\text{friction}} = 0.5 \times 84.87 \, \text{N} \approx 42.44 \, \text{N} \] ### Step 6: Calculate the total force required to pull the body up the incline 1. The total force (F) required to overcome both the gravitational component and the frictional force is given by: \[ F = W_{\text{parallel}} + F_{\text{friction}} \] \[ F = 49 \, \text{N} + 42.44 \, \text{N} \approx 91.44 \, \text{N} \] ### Conclusion The minimum force required to pull the body up the plane is approximately **91.44 N**. ---
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A body of mass 5 xx 10^(-3) kg is launched upon a rough inclined plane making an angle of 30^(@) with the horizontal Obtain the coefficient of friction between the body and the plane if the time of ascent is half the time of descent .

A body of mass 5 xx 10^(-3) kg is launched upon a rough inclined plane making an angle of 30° with the horizontal. Find the coefficient of friction between the body and the plane if the time of ascent is half of the time of descent.

Knowledge Check

  • In the above problem, the minimum force required to pull the body up the inclined plane

    A
    66.6 N
    B
    86.6 N
    C
    96.6 N
    D
    76.6 N
  • A body of mass 5xx10^(-3) kg is launched upon a rough inclined plane making an angle of 30^(@) with the horizontal. Obtain the coefficient of friction between the body and the plane if the time of ascent is half of the time of descent.

    A
    0.346
    B
    0.921
    C
    1.926
    D
    2.912
  • A body is sliding down an inclined plane forming an angle 30^(@) with the horizontal. If the coefficient of friction is 0.3 then acceleration of the body is

    A
    `1.25 ms^(-2)`
    B
    `2.35 ms^(-2)`
    C
    `3.4 ms^(-2)`
    D
    `4.9 ms^(-2)`
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