a) Normal force in the 1st case: 18.9 N, in the 2nd case: 6.7 N
b) The weight of the two balls is 19.6 N
c) The net force along the ramp is: 5.1 N (1st ball), 18.4 N (2nd ball)
Explanation:
a)
The free-body diagram for the two ramps is shown in attachment. Note that the diagram for the two balls is exactly the same, the only difference is the magnitude of the angle of the ramp, [tex]\theta[/tex] (15 degrees in one case, 70 degrees in the other case).
In order to calculate the magnitude of the normal force (N), we have to study the forces acting in the direction perpendicular to the ramp. We have two forces:
Since the ball is in equilibrium in this direction, the net force is zero, so the equation of the forces is:
[tex]N-mg cos \theta = 0[/tex]
Where
m = 2 kg
[tex]g=9.8 m/s^2[/tex]
For the first ball, [tex]\theta=15^{\circ}[/tex], so the normal force is
[tex]N_1 = mg cos \theta = (2)(9.8)(cos 15^{\circ})=18.9 N[/tex]
For the second ball, [tex]\theta=70^{\circ}[/tex], so the normal force is
[tex]N_2 = mg cos \theta = (2)(9.8)(cos 70^{\circ})=6.7 N[/tex]
2)
The magnitude of the downward force acting on the ball, which is the force of gravity, is given by
[tex]F=mg[/tex]
where
m is the mass of the ball
g is the acceleration of gravity
We see that its value does not depend on the angle of the ramp, so it is the same for the two balls. We have:
m = 2 kg
[tex]g=9.8 m/s^2[/tex]
Therefore the weight of the two balls is
[tex]F=(2)(9.8)=19.6 N[/tex]
3)
In this problem, we have to analyze the forces acting in the direction parallel to the ramp.
We have only one force acting in this direction, down along the ramp: the component of the weight parallel to the ramp, which is given by
[tex]F=mg sin \theta[/tex]
m = 2 kg is the mass
[tex]g=9.8 m/s^2[/tex]
For the 1st ball, we have:
[tex]F=(2)(9.8)(sin 15^{\circ})=5.1 N[/tex]
For the 2nd ball, we have
[tex]F=(2)(9.8)(sin 70^{\circ})=18.4 N[/tex]
Learn more about ramps:
brainly.com/question/5884009
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