The coefficient of static friction between the 3.00 kg crate and the 35.0 degree incline is 0.300. What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
Added by Amanda B.
Step 1
First, we need to find the gravitational force acting on the crate. This can be calculated using the formula: F_gravity = m * g where m = 3.00 kg (mass of the crate) and g = 9.81 m/s^2 (acceleration due to gravity) Show more…
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The coefficient of static friction between the 3.00-kg crate and the $35.0^{\circ}$ incline of Figure $P 4.31$ is 0.300 , What minimum force $\overline{\mathbf{F}}$ must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
The coefficient of static friction between the $3.00 \mathrm{kg}$ crate and the $35.0^{\circ}$ incline shown here is $0.300 .$ What is the magnitude of the minimum force, $F$, that must be applied to the crate perpendicularly to the incline to prevent the crate from sliding down the incline?
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