The 49 kg climber is supported in the "chimney" by the friction forces exerted on his shoes and back. The climber has reduced push to verge of slip. The static coefficients of friction between his shoes and the wall, and between his back and the wall, are 1.2 and 0.8, respectively. What is the magnitude of the push he must exert? What fraction of the climbers weight is supported by the frictional force on his shoes?

force of push = (mg)/(u1+u2), so

(49*9.8)/(1.2+.8)=240.1N

fraction of weight= (u1)/(u1+u2),so
(1.2)/(1.2+.8)=.6

To calculate the magnitude of the push the climber must exert, we need to consider the forces acting on him.

1. Weight: The weight of the climber is given as 49 kg. The weight can be calculated by multiplying the mass (49 kg) by the acceleration due to gravity (9.8 m/s^2). So, weight = 49 kg x 9.8 m/s^2 = 480.2 N.

2. Friction force on shoes: The static coefficient of friction between the shoes and the wall is given as 1.2. The frictional force can be calculated by multiplying the static coefficient of friction by the weight. Friction force on shoes = 1.2 x 480.2 N = 576.24 N.

3. Friction force on the back: The static coefficient of friction between the back and the wall is given as 0.8. The frictional force can be calculated by multiplying the static coefficient of friction by the weight. Friction force on the back = 0.8 x 480.2 N = 384.16 N.

To calculate the magnitude of the push the climber must exert, we need to consider the equilibrium condition. Since the climber is on the verge of slipping, the push would balance the frictional forces.

Magnitude of the push = Friction force on shoes + Friction force on back
= 576.24 N + 384.16 N
= 960.4 N

Therefore, the magnitude of the push the climber must exert is 960.4 N.

To calculate the fraction of the climber's weight supported by the frictional force on his shoes, we can divide the friction force on shoes by the weight.

Fraction of weight supported by frictional force on shoes = (Friction force on shoes / Weight) x 100%
= (576.24 N / 480.2 N) x 100%
= 1.1994 x 100%
= 119.94%

Therefore, the fraction of the climber's weight supported by the frictional force on his shoes is approximately 119.94%.

To find the magnitude of the push the climber must exert, we need to consider the frictional forces acting on him.

The force required to prevent the climber from slipping down the wall can be calculated using the formula:

F = μN

Where:
F is the force of friction,
μ is the coefficient of friction, and
N is the normal force.

Let's calculate the forces:

1. Calculating the force of friction between the climber's shoes and the wall:
Given that the coefficient of friction (μ) between shoes and wall is 1.2, and the normal force (N) acting on the climber is equal to his weight (W), which is calculated as the mass (m) times the acceleration due to gravity (g), we can write:

F_shoes = μ_shoes * N = μ_shoes * W.

Substituting the values:
F_shoes = 1.2 * 49 kg * 9.8 m/s^2

2. Calculating the force of friction between the climber's back and the wall:
Given that the coefficient of friction (μ) between the back and the wall is 0.8, we can write:

F_back = μ_back * W.

Substituting the values:
F_back = 0.8 * 49 kg * 9.8 m/s^2.

Now, regarding the fraction of the climber's weight supported by the frictional force on his shoes, we can calculate it by dividing the force of friction on the shoes by the climber's weight:

Fraction = (F_shoes / W) * 100%.

Substituting the values:
Fraction = (F_shoes / (49 kg * 9.8 m/s^2)) * 100%.

Let's calculate the answers:

1. Calculate the magnitude of the push the climber must exert:
F_push = F_shoes + F_back.

2. Calculate the fraction of the climber's weight supported by the frictional force on his shoes:
Fraction = (F_shoes / (49 kg * 9.8 m/s^2)) * 100%.

Please note that I will provide the numerical answers after calculating the values above.

Well, if he is at the point of slip, then force friction shoes=X*mg

where x is some fraction of weight.
This means force normal=K, and mu*K= Xmg

Now if K is the normal force, then K must be acting on the back, so friction on the back is .8*mg*(1-x)

we know the two frictions have to equal weight,

.8(mg)(1-x)+1.2*X*mg=mg
.8-.8x+ 1.2x=1
.4x=.2
x= .5
so half his weight is supported by his shoes.