The box slides down the **wall **unless an external force of magnitude 23 N is applied on it. The **object** is directed upward with an angle of 27° above the** horizontal surface**. Therefore, the **mass** of **block** is 1.90 kg

<h3>What is friction?</h3>

A **friction** is a kind of force which resists the sliding or rolling of **objects** over the surface of each other.

**Applied force**, F = 23 N

Coefficient of static friction, μs = 0.40

Coefficient of kinetic friction, μs = 0.30

θ = 27°

Let 'N' be the normal reaction of the **wall **acting on the block and 'm' be the mass of **block**.

Resolve the components of force 'F'

As the block is in **horizontal equilibrium** with the wall.

So,

F Cos27° = N

N = 23 Cos27° = 20.495 N

As the block does not slide so it means that the** static friction **force acting on the **block** balances the **downwards forces** (gravity) acting on the block.

The force of **static friction** is:

μs x N = 0.4 x 20.495 = 8.19 N .... (1)

The** vertically downward force** acting on the block is (mg - F Sin27°)

mg - 23 Sin 27° = mg - 10.441 ... (2)

Now by **equating** the **forces** from equation (1) and (2), we get

mg - 10.441 = 8.19

mg = 18.631

m x 9.8 = 18.631

**m = 1.90 kg**

Thus, the **mass** of block is **1.90 kg**.

Learn more about** Friction** here:

brainly.com/question/13000653

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Your question is incomplete, most probably the complete question is:

A small box is held in place against a rough vertical wall by someone pushing on it with a force directed upward at 27 ∘ above the horizontal. The coefficients of static and kinetic friction between the box and wall are 0.40 and 0.30, respectively. The box slides down unless the applied force has magnitude 23 N. What is the mass of the box in kilograms?