Question

problem
given:
m: 80 kg
a: 0 m/s²
μₛ: 0.7
θ: 25°
wanted: applied force (fₐ) in n?

1. given:

fₐ: 390 n
fₓ: 900 n
a: 0 m/s²
wanted: coefficient of kinetic friction (μₖ)?

1. given:

m: 800 kg
a: 0 m/s²
μₖ: 0.5
wanted: force of friction (fₓ) in n?

1. given:

m: 8 kg
a: 0 m/s²
μₖ: 0.6
θ: 27°
wanted: applied force (fₐ) in n?

1. given:

fₓ: 1 n
m: 2.04 kg
a: 0 m/s²
θ: 28°
wanted: coefficient of kinetic friction (μₖ)?

Explanation:

1.

Step1: Analyze forces in y - direction

$N - mg\cos\theta= 0$, so $N = mg\cos\theta$.

Step2: Analyze forces in x - direction

$F_a-\mu_kN - mg\sin\theta=ma$. Since $a = 0$, $F_a=\mu_kmg\cos\theta+mg\sin\theta$.
Substitute $m = 80$ kg, $\mu_k = 0.7$, $\theta = 25^{\circ}$, $g = 9.8$ m/s².
$N=80\times9.8\times\cos(25^{\circ})\approx80\times9.8\times0.9063 = 709.77$ N.
$F_a=0.7\times709.77+80\times9.8\times\sin(25^{\circ})$
$F_a = 496.84+80\times9.8\times0.4226$
$F_a=496.84 + 331.27=828.11$ N

Step1: Analyze forces in x - direction

Since $a = 0$, $F_a - F_f=0$, so $F_f=F_a = 390$ N. Also, $F_f=\mu_kF_N$.

Step2: Solve for $\mu_k$

$\mu_k=\frac{F_f}{F_N}$. Substitute $F_f = 390$ N and $F_N = 900$ N.
$\mu_k=\frac{390}{900}\approx0.433$

Step1: Since $a = 0$, the force of friction balances the applied - like force (assuming no other horizontal forces).

The formula for the force of friction is $F_f=\mu_kN$. And since $a = 0$, $N = mg$.

Step2: Calculate $N$

$N=mg=800\times9.8 = 7840$ N.

Step3: Calculate $F_f$

$F_f=\mu_kN$. Substitute $\mu_k = 0.5$ and $N = 7840$ N.
$F_f=0.5\times7840 = 3920$ N

Answer:

$828.11$

2.