A car with a weight of 15,000N is being towed up a 20 degree slope at constant velocity. Friction is negligible. The tow rope is rated at 6000 N maximum tension. What are the forces acting on the car?
Question
A car with a weight of 15,000N is being towed up a 20 degree slope at constant velocity. Friction is negligible. The tow rope is rated at 6000 N maximum tension. What are the forces acting on the car?
Solution
The forces acting on the car are:
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Gravity: This is the weight of the car, which is given as 15,000N. This force acts vertically downwards.
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Normal Force: This is the force exerted by the surface (in this case, the slope) on the car. It acts perpendicular to the surface. Since the car is on a slope, the normal force is less than the weight of the car. It can be calculated using the formula: Normal Force = Weight * cos(θ), where θ is the angle of the slope. So, Normal Force = 15,000N * cos(20°).
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Tension: This is the force exerted by the tow rope on the car. It acts along the direction of the rope, pulling the car up the slope. The maximum tension the rope can handle is given as 6000N.
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Friction: It is given that friction is negligible, so we can ignore this force in this case.
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The force due to the slope: This is the component of the car's weight that acts parallel to the slope, trying to pull the car down the slope. It can be calculated using the formula: Force due to slope = Weight * sin(θ), where θ is the angle of the slope. So, Force due to slope = 15,000N * sin(20°).
So, the forces acting on the car are the tension in the rope, the normal force, the force due to the slope, and gravity.
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