Question

1. As the angle of the ramp is increased, the normal force increases /decreases / remains the same and the friction-force increases /decreases / remains the same. [1 Point] 2. As the angle of the ramp is increased, the force parallel increases /decreases / remains the same. [1 Point] 3. The angle at which the force down the plane was equal to the force of friction (for the cabinet) was _____________. [1 Point] 4. Consider a very low (~ zero) friction, 5.0 kg skateboard on a ramp at an angle of 15o to the horizontal. What would be the net force that would cause acceleration when the skateboard is allowed to move

Answer 1

(1) As the angle of the ramp is increased, the normal force decreases.

(2) As the angle of the ramp is increased, the parallel force increases.

(3) The angle at which the force down the plane was equal to the force of friction is zero degree.

(4) The net force that would cause acceleration is 47.33 N.

Let the angle of inclination of the ramp = θ

(1)

The normal force on an object on the ramp inclined to the ramp is calculated as follows;

$F_n = mgcos (\theta)$

when θ is 0;

$F_n = mgcos (0)\\\\F_n = mg$

when θ is 90;

$F_n = mgcos(90)\\\\F_n = 0$

Thus, as the angle of the ramp is increased, the normal force decreases.

(2)

The parallel force on an object on the ramp inclined to the ramp is calculated as follows;

$F_x = mgsin(\theta)$

when θ is 0;

$F_x = mgsin(\theta)\\\\F_x = mgsin(0) \\\\F_x = 0$

when θ is 90;

$F_x = mgsin(90)\\\\F_x = mg$

Thus, as the angle of the ramp is increased, the parallel force increases.

(3)

The force of friction is calculated as follows;

$F_n = \mu F_n$

$F_k = \mu mgcos(\theta)$

$F_k = \mu mg cos(0)\\\\F_k = \mu mg$

Thus, the angle is zero degree

(4)

The net force that would cause acceleration is calculated as follows;

$F_k = Fn\\\\F_k = mg cos(\theta)\\\\F_k = 5 \times 9.8 \times cos(15)\\\\F_k = 47.33 \ N$

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