Contact force of block on a wall

[dodosenpai]

Homework Statement

In this sketch, the mass m = 1.6 kg is pushed against a wall by force F, as shown, in the horizontal direction. The values of the coefficients of kinetic and static friction for the contact between are 0.81 and 0.84 respectively. The magnitude of F is slowly and continuously reduced. Just before the block falls, what is the total contact force Fc exerted by the block on the wall. What is the magnitude of that force and what is the angle θ it makes with horizontal? (+ve for above, -ve for below the horizontal).

Homework Equations

F_f = μ_sN

The Attempt at a Solution

Since the block is not moving:
Input F = N
F_f = F_g

Therefore mg = μ_s*N

Rearrange we get

N = mg/μ_s
= (1.6*9.8)/0.84
= 18.66667
= 19(2 sig. fig.)

Since Input F = N, wouldn't the contact force of the mass onto the wall be 19N? Apparently this is wrong.

 

[vela]

There are two forces the wall exerts on the block. The problem is asking for the sum of the normal force and the force of static friction.

 

[dodosenpai]

So that would be that the contact force is not horizontal, but instead has a magnitude of √((0.84*19)²+(19)²) = 24.81 = 25N at an angle of tan^-1(0.84) = 40.03 = 40°?

 

[dodosenpai]

There are two forces the wall exerts on the block. The problem is asking for the sum of the normal force and the force of static friction.

So that would be that the contact force is not horizontal, but instead has a magnitude of √((0.84*19)2+(19)2) = 24.81 = 25N at an angle of tan-1(0.84) = 40.03 = 40°?

 

[vela]

Yup, but I wouldn't use the rounded value for N in your calculations. Round to the correct number of sig figs only at the end of the calculation.

 

[dodosenpai]

Thanks