Question

Chapter 06, Problem 005 A 2.30 kg block is initially at rest on a horizontal surface. A horizontal force F of magnitude 4.82 N and a vertical force P are then ap block (see the figure). The coefficients of friction for the block and surface are us-04 and μ.-0.25. Determine the magnitude of the frictional force acting on the block if the magnitude of Pis (a)9.00 N and (b)12.0 N. (The upward pull is insufficient to move the block vertically.) plied to the (a) Number Units 0 (b) Number Units

Answer 1

here block initially at rest on the horizontal surface

mass , m = 2.5 kg , μ_s = 0.40 , μ_k = 0.25 , F = 4.82 N

if draw force diagram then ,

here if Fs ≥  4.82 N , then F_s is equal to 4.82 N, if F_s < 4.82 N then block moves and we have to find kinetic friction,

F_k = μ_{k N,}

here from force diagram , initially block is at rest , so net force should be zero. so we take y component of force is equal to zero.

$\textstyle\sum$ F_y = m a_y = 0

from diagram, we can write sum of forces according to their direction,

N + P = m g =0 ( here a_y = g )

N = mg - P

N = 2.30 * 9.8 - P ( m = 2.3 kg , g = 9.8 m/s² )

N = 22.54 - P

we use this equation for finding normal force N, and then we put value of N in the static friction force equation F_s = μ_s N

here for , case -(a)

here vertical force , P = 9 N , So we put this value in the equation,

N = 22.54 - P

N= 22.54 - 9

N = 13.54 N

so here normal force N = 13.54 N,

static friction force, F_s = μ_s N

= 0.40 * 13.54

F_s = 5.416 N

Here static friction force F_s = 5.416 N is greater then horizontal force F,

F_s > F ( 5.416> 4.82)

Then here F_s = 4.82 N

the static friction is more then force acting on body . so body at rest. and friction force direction is to left side.

(case-b)

here vertical force , P = 12 N , So we put this value in the equation,

N = 22.54 - P

N= 22.54 - 12 ( P =12 N)

N = 10.54 N

so here normal force N = 10.54 N,

static friction force, F_s = μ_s N

= 0.40 * 10.54

F_s = 4.216 N

Here static friction force F_s = 4.216 N is less then horizontal force F,

F_s< F (4.216 < 4.82)

here static friction force is less then the horizontal force F

so here the block start moving when vertical force P = 12 N applied , so we have to find kinetic friction force for the block,

here kinetic friction force, F_k = μ_k N

Here for this case we have value of N = 10.54 N and μ_k = 0.25

F_k = 0.25 * 10.54

  F_k = 2.63 N

so here when we applied vertical force P 12 N ,then block start moving. so we are find kinetic friction force for the block ,    F_k = 2.63 N

kinetic friction force is in direction to the left