Question

find the tension on the rope that will keep 3200 lb car on a 30 degrees incline when athe rope is parallel to the incline; when the rope makes 60 degrees. With the horizontal .Neglect fiction.

Answer 1

According to the Newton's first law: $\Sigma\vec{F}=0$ and

$\Sigma\vec{F_{x}}=0$ , where $\vec{F_{x}}$- projection of any force to the x axis.

1) When the rope is parallel to the incline:

$|\vec F|= |\vec G_{x}|=|\vec G|\cdot sin(30°)=3200\cdot\frac{1}{2}=1600$ lb.

2) When the rope makes 60 degrees with the horizontal:

$|\vec F_{x}|= |\vec G_{x}|=|\vec G|\cdot sin(30°)=3200\cdot\frac{1}{2}=1600$ lb;

$|\vec F|=\frac{ |\vec F_{x}|}{cos(30°)}=\frac{1600}{\sqrt3/2}\approx1847.5$ lb.

Answer: 1)when the rope is parallel to the incline the tension on the rope is 1600 lb;

2)when the rope makes 60 degrees with the horizontal the tension on the rope is about 1847.5 lb.