Respuesta :
Let a be the linear acceleration of the center of mass, fs the magnitued of the frictional force, I be the rotational inertia of the ball, M be the mass, R the moment arm, and g be the gravity.
For any body rolling along an incline of angle O with the horizontal, a is :
a = - g sin O / ( 1 + (I / MR^2))
and
fs = - I (a / R^2)
Thus, substituting the given value to the formulas above, we get (assuming g = 9.8 m/s^2 and I = 2/5 MR^2 for a solid sphere)
a = -(9.8 m/s^2)sin30/(1 + ((2/5MR^2) / (MR^2)))
= -(9.8 m/s^2) sin 30/ (1 + (7/5))
= -3.5 m/s^2
and
fs = - (2/5MR^2)a / R^2
= - (2/5M)a
= - (2 * 8 / 5) ( -3 .5)
= 11.2 N
For any body rolling along an incline of angle O with the horizontal, a is :
a = - g sin O / ( 1 + (I / MR^2))
and
fs = - I (a / R^2)
Thus, substituting the given value to the formulas above, we get (assuming g = 9.8 m/s^2 and I = 2/5 MR^2 for a solid sphere)
a = -(9.8 m/s^2)sin30/(1 + ((2/5MR^2) / (MR^2)))
= -(9.8 m/s^2) sin 30/ (1 + (7/5))
= -3.5 m/s^2
and
fs = - (2/5MR^2)a / R^2
= - (2/5M)a
= - (2 * 8 / 5) ( -3 .5)
= 11.2 N
The magnitude of the frictional force on the sphere will be 11.2 N . Friction force is the opposition force.
What is friction force?
It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically it is defined as the product of the coefficient of friction and normal reaction.
The given data in the problem is;
u is the initial velocity= 0
[tex]\rm \theta[/tex] is the inclined plane angle = 30°
m is the mass of the sphere = 8 kg
r is the radius = 0.19 m
[tex]\mu_s[/tex] is the coefficient of static friction = 0.64
F is the magnitude of the frictional force =?
The acceleration of the rolling up an inclination of angle with the horizontal;
[tex]\rm a = - \frac{g sin \theta}{( 1 + (I / MR^2))} \\\\ \rm a= \frac{9.81 \times sin 30^0 \times MR^2}{\frac{2}{5} MR^2} \\\\ \rm a = - 3.5 \ m/sec^2\neq[/tex]
The friction force will be;
[tex]\rm f_s = \frac{\frac{2}{5}MR^2 \times a }{R^2} \\\\ \rm f_s =\frac{2}{5} Ma \\\\ \rm f_s =2 \times \frac{8}{5} \times (-3.5) \\\\ \rm f_s =11.2 N[/tex]
Hence the magnitude of the frictional force on the sphere will be 11.2 N
To learn more about the friction force refer to the link;https://brainly.com/question/1714663
