arch 207 (exam 1)\nsolve the following problems. write your final answer in the space provided. good…

Explanation:

Step1: Identify the problem type

This is a statics - friction and equilibrium problem in engineering mechanics.

Step2: Situation 1 - Calculate normal force on 50 - kg crate

The weight of the 50 - kg crate is $W =mg$, where $m = 50$ kg and $g=9.81$ m/s², so $W=50\times9.81 = 490.5$ N. The force $P$ has components. If $P$ is applied at an angle, assume $P$ has a horizontal component $P_x$ and a vertical component $P_y$. For a force $P$ with direction ratios 3 - 4 - 5, if $P = 200$ N, $P_x=\frac{4}{5}P = 160$ N and $P_y=\frac{3}{5}P = 120$ N. The normal force $N$ on the crate is $N=W - P_y=490.5-120 = 370.5$ N. The maximum static - friction force $F_{s,max}=\mu_sN=0.3\times370.5 = 111.15$ N. Since $P_x = 160$ N>$F_{s,max}$, the crate is moving and the kinetic - friction force $F_k=\mu_kN$. Assuming $\mu_k=\mu_s = 0.3$, $F_k = 0.3\times370.5=111.15$ N. When $P = 400$ N, $P_x=\frac{4}{5}\times400 = 320$ N and $P_y=\frac{3}{5}\times400 = 240$ N. The normal force $N=W - P_y=490.5 - 240=250.5$ N. The maximum static - friction force $F_{s,max}=\mu_sN=0.3\times250.5 = 75.15$ N. Since $P_x = 320$ N>$F_{s,max}$, the crate is moving and $F_k=\mu_kN=0.3\times250.5 = 75.15$ N.

Step3: Situation 2 - Calculate normal force

(a)

For the 50 - N block with a 20 - N force at 55°: The vertical component of the 20 - N force is $F_y = 20\sin55^{\circ}\approx16.383$ N. The normal force $N = 50 - 16.383=33.617$ N.

(b)

For the 60 - N block on a 40° incline: The normal force $N = 60\cos40^{\circ}\approx45.963$ N.

(c)

For the 60 - N block with a 70 - N force at 40° on a 40° incline: The normal - force component due to the weight is $N_1 = 60\cos40^{\circ}\approx45.963$ N. The vertical component of the 70 - N force is $F_y=70\sin40^{\circ}\approx44.995$ N. The normal force $N = 60\cos40^{\circ}+70\sin40^{\circ}\approx45.963 + 44.995=90.958$ N.

Step4: Situation 3 - Equilibrium on incline

For the 200 - N block on a 35.0° incline, the component of the 200 - N block's weight along the incline is $F_{parallel}=200\sin35.0^{\circ}\approx114.715$ N. For the block to remain at rest, the weight $F_W$ of the object on the right must balance this component, so $F_W = 114.715$ N.

Step5: Situation 4 - Maximum force without movement

The weight of the 250 - lb crate is $W = 250$ lb. The maximum static - friction force $F_{s,max}=\mu_sN$. The normal force $N = W=250$ lb. So $F_{s,max}=0.4\times250 = 100$ lb. The maximum force $P$ that can be applied without causing movement is equal to the maximum static - friction force, so $P = 100$ lb.

Step6: Situation 5 - Minimum force to prevent sliding

The weight of the 30 - kg rod $AB$ is $W=mg=30\times9.81 = 294.3$ N. Let's take moments about point $B$. The normal force at $A$ is $N_A$. The friction force at $A$ is $F_A=\mu_sN_A$. The length of the rod $L=\sqrt{3^{2}+4^{2}} = 5$ m. Taking moments about $B$: $W\times2 = N_A\times3+F_A\times4$. Since $F_A=\mu_sN_A = 0.2N_A$, we have $294.3\times2=N_A\times3 + 0.2N_A\times4$. $588.6=N_A(3 + 0.8)$. $N_A=\frac{588.6}{3.8}\approx154.895$ N. The minimum force $P = F_A=\mu_sN_A=0.2\times154.895 = 30.979$ N.

Answer:

Situation 1: a) $F = 111.150$ N, b) $F = 75.150$ N Situation 2: a) $N = 33.617$ N, b) $N = 45.963$ N, c) $N = 90.958$ N Situation 3: $F_W = 114.715$ N Situation 4: $P = 100$ lb Situation 5: $P = 30.979$ N