Question

A crate is supported by two ropes. One rope makes an angle of 46° 20′ with the horizontal and has a tension of 89.6 lb on it. The other rope is horizontal. Find the weight of the crate and the tension in the horizontal rope.

Accepted Answer

Identify the forces acting on the crate: the weight (W) acting vertically downward, the tension in the inclined rope ($$T_1 = 89.6 lb) at an $$angle of 46° 20′ with the horizontal, and the tension in the horizontal rope ($$T_2) $$acting horizontally. Resolve the tension in the inclined rope into its horizontal and vertical components using trigonometric functions: the horizontal component is $$T_1$$ \cos(46^\circ 20')$$ and the vertical component is T_1$ \sin(46^\circ 20'). $$Since the crate is in equilibrium, apply the condition that the sum of forces in the horizontal direction is zero: $$T_2$$ = $$T_1$$ \cos(46^\circ 20')$$. Apply the equilibrium condition in the vertical direction: the upward vertical component of the inclined rope's tension balances the weight of the crate, so W$$ = $$T_1$$ \sin(46^\circ 20')$$. Use the above equations to express the weight of the crate and the tension in the horizontal rope in terms of the given tension and angle, then calculate their numerical values if needed.

A crate is supported by two ropes. One rope makes an angle of 46° 20′ with the horizontal and has a tension of 89.6 lb on it. The other rope is horizontal. Find the weight of the crate and the tension in the horizontal rope.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Resolving Forces into Components

Equilibrium of Forces

Trigonometric Functions and Angle Conversion