A toboggan of mass 18.0 kg has gotten away and is sliding down the hillside which makes a incline of 33.0 degrees to the horizontal. If the toboggan's acceleration is 0.300 m/s^2, how large is the coefficient of kinetic friction?
Please show me how to do this.
Please show me how to do this.
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start by drawing a diagram of an object on an incline of 33 deg
the weight vector is straight down and can be resolved into a component perpendicular to the incline (Fn --- the normal force) and a component parallel to the surface of the incline (Fp).
Fn and Fp are at a right angle and form right triangles with Fw as the hypotenuse.
In this case
Fw = m g = 18.0 kg (9.81 m/s/s) = 176.6 N
then from the diagram (after you chase some angles down)
Fn = Fw cos33 = 176.6 (0.839) = 148.1 N
Fp = Fw sin33 = 176.6 (0.545) = 96.18 N
and
Ff (force of friction) = mu Fn
and you are looking for mu
Fnet = m a
and
the acceleration is parallel to the surface of the incline so
Fnet = Fp - Ff = Fp - mu Fn
giving
m a = Fp - mu Fn
18.0 (0.300) = 96.18 - 148.1 mu
5.40 = 96.18 - 148.1 mu
148.1 mu = 96.18 - 5.40 = 90.78
mu = 0.613 (to 3 s.f.)
the weight vector is straight down and can be resolved into a component perpendicular to the incline (Fn --- the normal force) and a component parallel to the surface of the incline (Fp).
Fn and Fp are at a right angle and form right triangles with Fw as the hypotenuse.
In this case
Fw = m g = 18.0 kg (9.81 m/s/s) = 176.6 N
then from the diagram (after you chase some angles down)
Fn = Fw cos33 = 176.6 (0.839) = 148.1 N
Fp = Fw sin33 = 176.6 (0.545) = 96.18 N
and
Ff (force of friction) = mu Fn
and you are looking for mu
Fnet = m a
and
the acceleration is parallel to the surface of the incline so
Fnet = Fp - Ff = Fp - mu Fn
giving
m a = Fp - mu Fn
18.0 (0.300) = 96.18 - 148.1 mu
5.40 = 96.18 - 148.1 mu
148.1 mu = 96.18 - 5.40 = 90.78
mu = 0.613 (to 3 s.f.)