You throw a Frisbee of mass m and radius r so that it is spinning about a horizontal axis perpendicular to the plane of the Fris

bee. Ignoring air resistance, the torque exerted about its center of mass by gravity is

You might be interested in

Sav [3153]

Answer: $F_{v} =\mu_{k} mg$

The force required has a magnitude of 2601.9 N

Explanation:

m = 450 kg

Static friction coefficient μs = 0.73

Kinetic friction coefficient μk = 0.59

The force necessary to initiate movement of the crate is $F_{s} =\mu_{s} mg$ .

Once the crate begins to move, the frictional force decreases to $F_{v} =\mu_{k} mg$ .

To maintain the motion of the crate at a steady velocity, we must lower the pushing force to $F_{v} =\mu_{k} mg$ .

Subsequently, the pushing force aligns with the frictional force stemming from kinetic friction, enabling balanced forces and consistent velocity.

<pMagnitude of the force

$F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N$

4 0

3 months ago