"The track is now lowered to a point where the cart just remains at rest while on the incline. If the track makes an angle of 7 degrees with the horizontal, what is the coefficient of static friction between the cart and the track?"

Respuesta :

Answer:

0.12

Explanation:

The coefficient of static force is the frictional force between two objects which are static.

The net force on the cart is the component of its gravitational force (weight) minus the frictional force. Therefore to prevent the cart from sliding down the track, the frictional force and gravitational force must be equal.

frictional force = gravitational force

[tex]\mu* mgcos\theta=mgsin\theta\\\\m=mass,g=acceleration\ due\ to\ gravity,\mu=coefficient\ of\ static\ friction,\theta=angle\ with\ horizontal\\\\\mu* mgcos\theta=mgsin\theta\\\\\mu=\frac{mgsin\theta}{mgcos\theta} \\\\\mu=tan\theta=tan(7\\\\\mu=0.12[/tex]

The coefficient of static friction between the cart and the track is 0.12.

Net force on the cart

The net force on the cart is calculated as follows;

∑F = 0 (since the cart is at rest, the net force will be zero)

F - Ff = 0

mgsinθ - μmgcosθ = 0

mgsinθ = μmgcosθ

sinθ = μcosθ

μ = sinθ/cosθ

μ = tanθ

μ = tan(7)

μ = 0.12

Thus, the coefficient of static friction between the cart and the track is 0.12.

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