situation 2. 13. find the normal force acting on the block in each of the equilibrium situations shown in…

Answer:

(a) $37.04\ N$; (b) $46.0\ N$; (c) $94.9\ N$

Explanation:

Step1: Analyze situation (a)

The weight of the block is $W = 50\ N$. The vertical - component of the applied force $F = 20\ N$ is $F_y=F\sin55^{\circ}$. The normal force $N$ is given by $N = W - F\sin55^{\circ}$. Substitute $F = 20\ N$ and $W = 50\ N$ into the formula: $N=50 - 20\times\sin55^{\circ}=50 - 20\times0.8192\approx37.04\ N$.

Step2: Analyze situation (b)

The weight of the block is $W = 60\ N$. Resolve the weight into components perpendicular and parallel to the inclined plane. The normal force $N$ on the inclined - plane is the component of the weight perpendicular to the plane. So $N = W\cos40^{\circ}$. Substitute $W = 60\ N$ into the formula: $N = 60\times\cos40^{\circ}=60\times0.766\approx46.0\ N$.

Step3: Analyze situation (c)

The weight of the block is $W = 60\ N$. The applied force is $F = 70\ N$. Resolve the weight and the applied force into components perpendicular to the inclined plane. The normal force $N$ is $N=W\cos40^{\circ}+F\sin40^{\circ}$. Substitute $W = 60\ N$ and $F = 70\ N$ into the formula: $N = 60\times\cos40^{\circ}+70\times\sin40^{\circ}=60\times0.766 + 70\times0.6428=45.96+45.0=94.9\ N$.