1. ## horizontal forces.

A man pulls a sledge of mass 90kg on horizontal ground by means of a rope inclined at 30o to the horizontal. The acceleration is 0.15m/s2 and the resistance to motion is 100N.

Determine
(i) the tension in the rope.
(ii) the normal reaction between sledge and ground.

2. Originally Posted by abz
A man pulls a sledge of mass 90kg on horizontal ground by means of a rope inclined at 30o to the horizontal. The acceleration is 0.15m/s2 and the resistance to motion is 100N.

Determine
(i) the tension in the rope.
(ii) the normal reaction between sledge and ground.
You have to resolve in component. This is moving horizontally hence you will equate it to $\displaystyle ma$ (Newton's Law). It is not moving vertically so you do not equate to $\displaystyle ma$ as acceleration will be $\displaystyle 0$ vertically. Drawing a diagram for these questions solves half the question.

(i) Resolve Horizontally: $\displaystyle R(\leftrightarrow): T\cos 30^{\circ} - 100 = 90(0.15)$. Solve for $\displaystyle T$

(ii) Resolve Vertically: $\displaystyle R(\updownarrow): R + T\sin 30^{\circ} - 90g = 0$. Solve for $\displaystyle R$.