A 25.0 Kg sled with a rope attached is being pulled down a slope with a 21 degree angle. There is an 8 Kg box on the sled.The tension gradually increases. At what value of tension does the box slip?
You need to know the coefficient of friction.
To find the value of tension at which the box starts to slip, we need to consider the forces acting on the box on the sled. The two main forces are the gravitational force (also known as weight) and the tension force from the rope.
Let's break down the forces acting on the box on the sled:
1. Gravitational force: The weight of the box is given by mass multiplied by acceleration due to gravity. The acceleration due to gravity is approximately 9.8 m/s^2.
Weight of the box = mass × acceleration due to gravity
Weight of the box = 8 kg × 9.8 m/s^2
2. Force parallel to the slope: The component of weight acting parallel to the slope is given by the weight of the box multiplied by sine of the angle of the slope.
Component of weight parallel to the slope = Weight of the box × sin(angle of slope)
Component of weight parallel to the slope = (8 kg × 9.8 m/s^2) × sin(21 degrees)
3. Tension force: The tension force in the rope is opposing the component of weight parallel to the slope. So, the maximum tension force at which the box will not slip is equal to the component of weight parallel to the slope.
Tension force = Component of weight parallel to the slope
Tension force = (8 kg × 9.8 m/s^2) × sin(21 degrees)
Calculating this value will give you the maximum tension at which the box on the sled will not slip.