Question

for acceleration due to gravity, use (a_{g}=9.81 m/s^{2}); ignore air - resistance unless the problem says otherwise. please try to get within 2% of the correct answer. you can enter your answers in decimal notation, or you can enter them in scientific notation using e or ×10^ for the power of 10 (e.g., enter (6.02\times10^{23}) as 6.02e23 or 6.02x10^23). problem answer 1 given: (f_{x}=2.00 n), (a = 0 m/s^{2}), (mu_{s}=0.3) wanted: normal force ((f_{n})) in n? normal force? n 2 given: (m = 4 kg), (a = 0 m/s^{2}), (mu_{s}=0.8), (\theta = 28^{circ}) wanted: applied force ((f_{a})) in n? applied force? n 3 given: (f_{a}=3000 n), (a = 0 m/s^{2}), (mu_{s}=0.8) wanted: normal force ((f_{n})) in n? normal force? n 4 given: (m = 6 kg), (a = 0 m/s^{2}), (mu_{k}=0.1) wanted: force of friction ((f_{f})) in n? force of friction? n 5 given: (m = 40.0 kg), (a = 0 m/s^{2}), (mu_{s}=0.4), (\theta = 10^{circ}) wanted: applied force ((f_{a})) in n?

Explanation:

Step1: Analyze problem 1

The object is in equilibrium ($a = 0\ m/s^{2}$). The normal force $F_N$ balances the other vertical - acting forces. Since no other vertical forces are mentioned except the given force $F = 2.00\ N$, by Newton's first law ($\sum F_y=0$), the normal force $F_N = 2.00\ N$.

Step2: Analyze problem 2

The object is in equilibrium ($a = 0\ m/s^{2}$). Resolve the forces in the vertical and horizontal directions. Let the applied force be $F_a$. In the vertical direction, $F_N=mg + F_a\sin\theta$. In the horizontal direction, $\mu_sF_N=F_a\cos\theta$. Substitute $F_N$ from the first equation into the second: $\mu_s(mg + F_a\sin\theta)=F_a\cos\theta$. Given $m = 4\ kg$, $g = 9.81\ m/s^{2}$, $\mu_s=0.8$, $\theta = 28^{\circ}$.

$$0.8\times(4\times9.81+F_a\sin28^{\circ})=F_a\cos28^{\circ}$$

$$0.8\times4\times9.81+0.8F_a\sin28^{\circ}=F_a\cos28^{\circ}$$

$$31.392=F_a(\cos28^{\circ}-0.8\sin28^{\circ})$$

$$F_a=\frac{31.392}{\cos28^{\circ}-0.8\sin28^{\circ}}\approx\frac{31.392}{0.8829 - 0.8\times0.4695}\approx\frac{31.392}{0.8829 - 0.3756}\approx\frac{31.392}{0.5073}\approx62\ N$$

Step3: Analyze problem 3

The object has $a = 0\ m/s^{2}$, so it is in equilibrium. If a force $F_a = 3000\ N$ is applied and there is no acceleration, the normal force $F_N$ balances the vertical component of the applied - force (assuming no other vertical forces). Since there is no information about the angle of the applied force and it seems to be a vertical - acting force, $F_N = 3000\ N$.

Step4: Analyze problem 4

The normal force $F_N=mg$ (since $a = 0\ m/s^{2}$), with $m = 8\ kg$ and $g = 9.81\ m/s^{2}$, so $F_N=8\times9.81 = 78.48\ N$. The force of friction $F_f=\mu_kF_N$, with $\mu_k = 0.1$. Then $F_f=0.1\times78.48 = 7.848\ N$.

Step5: Analyze problem 5

Similar to problem 2, in the vertical direction $F_N=mg + F_a\sin\theta$, in the horizontal direction $\mu_sF_N=F_a\cos\theta$. Substitute $F_N$ into the second equation: $\mu_s(mg + F_a\sin\theta)=F_a\cos\theta$. Given $m = 40.0\ kg$, $g = 9.81\ m/s^{2}$, $\mu_s = 0.4$, $\theta = 10^{\circ}$.

$$0.4\times(40\times9.81+F_a\sin10^{\circ})=F_a\cos10^{\circ}$$

$$0.4\times40\times9.81+0.4F_a\sin10^{\circ}=F_a\cos10^{\circ}$$

$$156.96=F_a(\cos10^{\circ}-0.4\sin10^{\circ})$$

$$F_a=\frac{156.96}{\cos10^{\circ}-0.4\sin10^{\circ}}\approx\frac{156.96}{0.9848-0.4\times0.1736}\approx\frac{156.96}{0.9848 - 0.0694}\approx\frac{156.96}{0.9154}\approx171\ N$$

Answer:

1. $2.00$
2. $62$
3. $3000$
4. $7.848$
5. $171$