Calculating Frictional Force on Inclined Block | 43 kg Mass | 19° Inclination

[grouchy]

a block is at rest on the incline shown in the figure. the coefficients of static and kinetic friction are s = 0.42 and k = 0.36 respectively. The acceleration of gravity is 9.8 m/s^2. the angle of inclination is 19 degrees.

what is the frictional force acting on the 43 kg mass? answer in units of N.


Honestly I have no idea what to do on this problem...

 

[PhanthomJay]

a block is at rest on the incline shown in the figure. the coefficients of static and kinetic friction are s = 0.42 and k = 0.36 respectively. The acceleration of gravity is 9.8 m/s^2. the angle of inclination is 19 degrees.

what is the frictional force acting on the 43 kg mass? answer in units of N.


Honestly I have no idea what to do on this problem...

If the block is at rest, you need only to identify all the forces acting on the block and apply Newton 1 to solve for the unknown friction force.

 

[ogb p]

I would approach this problem by first understanding the concept of frictional force and how it relates to the given scenario. Frictional force is a resistive force that opposes the motion of an object and is dependent on the coefficient of friction, the normal force, and the angle of inclination.

In this case, the block is at rest on the incline, meaning the force of gravity pulling it down the incline is equal to the force of friction pushing it up the incline. The coefficient of static friction (s) is used when the object is at rest, while the coefficient of kinetic friction (k) is used when the object is in motion.

To calculate the frictional force in this scenario, we can use the equation Ffr = μFn, where Ffr is the frictional force, μ is the coefficient of friction, and Fn is the normal force. The normal force is equal to the component of the force of gravity perpendicular to the incline, which can be calculated using the angle of inclination and the mass of the block.

Substituting the given values, we get Ffr = (0.42)(43 kg)(9.8 m/s^2)(cos 19°) = 168.7 N. This is the maximum force of friction that can act on the block before it starts to slide down the incline. Since the block is at rest, the actual frictional force acting on it is equal to the maximum force of static friction, which is 168.7 N in this case.

In conclusion, the frictional force acting on the 43 kg mass on the 19° inclined block is 168.7 N.

 

[m2003]

I would start by identifying the relevant equations and factors involved in calculating frictional force. In this case, we can use the equation Ff = μN, where Ff is the frictional force, μ is the coefficient of friction, and N is the normal force. The normal force is the force exerted by the incline on the block, which can be calculated using the equation N = mgcosθ, where m is the mass of the block, g is the acceleration of gravity, and θ is the angle of inclination.

Plugging in the given values, we have N = (43 kg)(9.8 m/s^2)cos19° = 407.78 N. Now we can calculate the frictional force by multiplying the normal force by the coefficient of static friction, since the block is at rest. This gives us Ff = (0.42)(407.78 N) = 171.38 N.

If we wanted to calculate the kinetic frictional force, we would use the same equation but with the coefficient of kinetic friction instead, since the block is in motion. This would give us Ff = (0.36)(407.78 N) = 146.80 N.

In summary, the frictional force acting on the 43 kg mass on the 19° inclined block is 171.38 N when the block is at rest and 146.80 N when the block is in motion.