Question

A 10.0kg mass is placed on a 25 degree incline and friction keeps it from sliding. The coefficient of static friction in this case is 0.580, and the coefficient of sliding friction is 0.520. What is the frictional force in this situation?

Please show your work. Thanks!

Answer 1

Answer 2

The forces acting on the block of mass 10 kg are the force due to gravity mg, normal force n and the force of static friction f_s. These forces balancing when the block is on the verge of slipping but not has started to move. Considering x to be parallel to the plane and y to be in its prependicular direction. For this case block is at rest. Using Newton's second law

$\sum F_x = mg sin \theta -f_s = m a_x=0 .....(1)$

$\sum F_y= n-mg cos \theta = m a_y=0 .....(2)$

From Eq. (1) , the force of static friction can be calculated as

$f_s= mg sin \theta = 10 kg \times9.8 m/s^2 \times sin25^0 = 41.4 N$

From Eq. (2) the normal force n

$n= mgcos \theta = 10 kg \times9.8 m/s^2 \times cos25^0 = 88.8 N$

If the block just slips then

$f_{smax}= \mu_s n = 0.580 \times 88.8 N= 51.5 N$

Once the block starts to move it accelerates down the incline and the force of kinetic friction f_k is

$f_{k}= \mu_k n = 0.520 \times 88.8 N= 46.18 N$