A grinding wheel of radius 0.280 m rotating on a frictionless axle is brought to rest by applying a constant friction force tangential to its rim. The constant torque produced by this force is 75.3 N · m. Find the magnitude of the friction force.
(friction force) * (radius) = torque
Solve for the friction force, in Newtons.
75.3/0.28 = ___ N
Torque=r x friction
so...solve for friction for
torque is known:75.3 Nm
radius is known:0.280m
so Torque/radius=75.3Nm/0.280m
meters cancel out =269N
To find the magnitude of the friction force, we'll use the relationship between torque and friction force.
The torque produced by a force is given by the equation:
τ = rFsinθ
where τ is the torque, r is the radius, F is the force, and θ is the angle between the force and the radius vector.
In this case, the force is tangential to the rim of the grinding wheel, so the angle θ between the force and the radius vector is 90 degrees (since the force is perpendicular to the radius vector).
Plugging in the given values:
τ = 75.3 N · m,
r = 0.280 m,
θ = 90 degrees.
We can rearrange the equation to solve for the force F:
F = τ / (r · sinθ)
Now we can calculate the magnitude of the friction force:
F = 75.3 N · m / (0.280 m · sin(90 degrees))
The sine of 90 degrees is equal to 1, so the equation simplifies to:
F = 75.3 N · m / 0.280 m
Calculating this expression:
F ≈ 269.64 N
Therefore, the magnitude of the friction force is approximately 269.64 N.