Auniform 8.0 m long ladder of mass 15.0 kg leans against a frictionless wall. the coefficient of static friction between the ladder and the floor is 0.32. the reaction force from the wall is: (a) 45.0 n (b) 74.0 n (c) 47.0 n (d) 50.0 n

(c) F= 47 N

Explanation:

We apply Newton's first law for the equlibrio in x-y

ΣFx=0 Formula (1)

F-Ff=0

ΣFy=0 Formula (2)

Fn-W=0

Where:

F : The reaction force from the wall (N)

Ff : Friction force (N)

Fn : Normal force (N)

W :  ladder Weight (N)

Known Data:

m= 15 Kg : mass ladder

μs: 0.32 : coefficient of static friction between the ladder and the floor

g= 9.8 m/s²

Problem development

We calculate the weight of the ladder (W):

W= m*g = 15 kg *9.8 m/s² = 147 N

look at the free body diagram of the ladder in the attached graphic

We apply formula (2) to calculate the normal force (Fn):

ΣFy=0 Formula (2)

Fn-147 N =0

Fn = 147 N

We calculate the force of static friction between the ladder and the floor (Ff):

Ff= μs*Fn =  0.32* 147 N = 47 N

We apply formula (1) to calculate The reaction force from the wall (F)

ΣFx=0

F- 47 N = 0

F= 47 N

what are the options.

explanation:

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .