CONSTANT FORCES
5) A 10kg block is allowed to slide down a ramp with uk=0.15
a) What is the value of the frictional force opposing the blocks slide down the ramp?
5a solution) mgsinO --> (10kg)(9.8) sin20 =34N
mgcosO --> (10kg)(9.8)cos20=92N
Normal Force= 92N [N=mgcosO]
Equation f=u*N
f=(0.15)(92N)
f=13.8 N
Explanation: the normal force is the component that is perpendicular to the surface on which the object is moving and can be calculated by multiplying the mass of the object with the gravity value* the cos theta, the angle of the inclined surface which is measured from the horizontal.
Given the value of the uk as 0.15, we can plug in the normal force value and u in the equation to find the friction. f=u*N. and get 13.8N which is the opposing force. It opposes the pulling force.
The normal force is the force exerted upwards by the surface and is less than the weight force due to the 20 degrees incline the block is experiencing.
Force explanations:
N force arrow: is 92N and can be calculated by multiplying mgcos(theta) or (10kg*9.8m/s2)cos(20)
N can be calculated using mgcos(theta) because the x-component of the weight vector is mgcos(theta) based on geometry -because every vector can be broken down into its vertical and horizontal components. mgcos(theta) is normal to the surface -thus equal to the normal (92N). N can be calculated this way also because mgcos(theta) and N are in the same direction and the two are equal because there is no movement in that direction.
f-arrow: means that there is friction in the left direction-opposing the block. Friction can be calculated using f=u*N and in this problem has the value of 13.8N -which means that 13.8N are opposing the 34N force pulling the block down the ramp.
mgsin(theta): force with which the block moved down the ramp :34N. Vertical component of weight force (10kg*9.8m/s2)sin(20).
mgcos(theta): horizontal component of the weight force and it is equal to the N force (92N) as they are in the same direction and because there is no movement in that direction. Calculated as (10kg*9.8m/s2)cos(20.
-What does f=u*N mean?
Frictional force is proportional to the coefficient of friction times the Normal. Friction is equal to the coefficient of friction times the normal.
b) what is the acceleration of the block?
Solution: Ef=ma 33.5N-13.8N =10kg*a
a=1.97m/s2
b explanation) to find the acceleration, the formula ef=ma is used. All the forces in the line of the block are taken into account. These forces are friction (13.8N), and the pulling force. The pulling force is due to the 20degree incline and can be calculated by multiplying the mass of the object with the gravity value and the angle theta (mgsin theta). The force that opposes mg sin theta or 33.5N is friction. Taking into account friction, the net force is 19.7N. This is the component of the gravitational force along the inclination minus the frictional force. Using algebra, we can divide 19.7N/10kg=1.97m/s2. The block is accelerating down the incline at 1.97m/s^2.
a) What is the value of the frictional force opposing the blocks slide down the ramp?
5a solution) mgsinO --> (10kg)(9.8) sin20 =34N
mgcosO --> (10kg)(9.8)cos20=92N
Normal Force= 92N [N=mgcosO]
Equation f=u*N
f=(0.15)(92N)
f=13.8 N
Explanation: the normal force is the component that is perpendicular to the surface on which the object is moving and can be calculated by multiplying the mass of the object with the gravity value* the cos theta, the angle of the inclined surface which is measured from the horizontal.
Given the value of the uk as 0.15, we can plug in the normal force value and u in the equation to find the friction. f=u*N. and get 13.8N which is the opposing force. It opposes the pulling force.
The normal force is the force exerted upwards by the surface and is less than the weight force due to the 20 degrees incline the block is experiencing.
Force explanations:
N force arrow: is 92N and can be calculated by multiplying mgcos(theta) or (10kg*9.8m/s2)cos(20)
N can be calculated using mgcos(theta) because the x-component of the weight vector is mgcos(theta) based on geometry -because every vector can be broken down into its vertical and horizontal components. mgcos(theta) is normal to the surface -thus equal to the normal (92N). N can be calculated this way also because mgcos(theta) and N are in the same direction and the two are equal because there is no movement in that direction.
f-arrow: means that there is friction in the left direction-opposing the block. Friction can be calculated using f=u*N and in this problem has the value of 13.8N -which means that 13.8N are opposing the 34N force pulling the block down the ramp.
mgsin(theta): force with which the block moved down the ramp :34N. Vertical component of weight force (10kg*9.8m/s2)sin(20).
mgcos(theta): horizontal component of the weight force and it is equal to the N force (92N) as they are in the same direction and because there is no movement in that direction. Calculated as (10kg*9.8m/s2)cos(20.
-What does f=u*N mean?
Frictional force is proportional to the coefficient of friction times the Normal. Friction is equal to the coefficient of friction times the normal.
b) what is the acceleration of the block?
Solution: Ef=ma 33.5N-13.8N =10kg*a
a=1.97m/s2
b explanation) to find the acceleration, the formula ef=ma is used. All the forces in the line of the block are taken into account. These forces are friction (13.8N), and the pulling force. The pulling force is due to the 20degree incline and can be calculated by multiplying the mass of the object with the gravity value and the angle theta (mgsin theta). The force that opposes mg sin theta or 33.5N is friction. Taking into account friction, the net force is 19.7N. This is the component of the gravitational force along the inclination minus the frictional force. Using algebra, we can divide 19.7N/10kg=1.97m/s2. The block is accelerating down the incline at 1.97m/s^2.