A vehicle having a mass of 1.5tonnes is standing on an incline of 18degree to the horizontal.Determine the components of its weight parallel and perpendicular to the road surface.
Fp = 1.5*sin18 = Parallel Force(Wt.)
Fn=1.5*cos18 = Normal or perpendicular force(Wt.).
To determine the components of the weight of the vehicle parallel and perpendicular to the road surface on the inclined plane, we need to use trigonometry.
First, let's convert the mass of the vehicle from tonnes to kilograms. Since 1 tonne is equal to 1000 kilograms, the mass of the vehicle is 1.5 tonnes * 1000 kg/tonne = 1500 kg.
Now, let's draw a diagram to visualize the situation:
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In the diagram, the road surface is represented by the horizontal line, and the incline is represented by the diagonal line.
The weight of an object is given by the equation W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
To find the component of the weight parallel to the road surface, we need to calculate W_parallel. We can do this by using the equation W_parallel = W * sin(θ), where θ is the angle of the incline.
W_parallel = 1500 kg * 9.8 m/s^2 * sin(18 degrees)
= 1500 * 9.8 * sin(18 degrees)
≈ 4262 N
To find the component of the weight perpendicular to the road surface, we need to calculate W_perpendicular. We can do this by using the equation W_perpendicular = W * cos(θ), where θ is the angle of the incline.
W_perpendicular = 1500 kg * 9.8 m/s^2 * cos(18 degrees)
= 1500 * 9.8 * cos(18 degrees)
≈ 14403 N
Therefore, the component of the weight of the vehicle parallel to the road surface is approximately 4262 N, and the component perpendicular to the road surface is approximately 14403 N.