A block weighing #6 kg# is on a plane with an incline of #(5pi)/6# and friction coefficient of #4/5#. How much force, if any, is necessary to keep the block from sliding down?
Julia Aguilar0.8N up the slope.
Start by setting your problem up in a diagram like so:
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Charlotte AaronThe force necessary to keep the block from sliding down the incline is equal to the component of the gravitational force parallel to the incline subtracted by the frictional force.
Gravitational force parallel to the incline = (weight of the block) * sin(θ) Frictional force = (coefficient of friction) * (normal force)
Weight of the block = mass * gravitational acceleration Normal force = weight of the block * cos(θ)
Substitute the given values and calculate:
Weight of the block = 6 kg * 9.8 m/s^2 Normal force = (weight of the block) * cos((5pi)/6)
Once you have these values, calculate the force needed to keep the block from sliding down by subtracting the frictional force from the gravitational force parallel to the incline.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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