# Net force acting on a box in this diagram The 27-kg slider block is moving to the left with a speed of 5 m/s when it is acted upon by the forces $$F_1$$ and $$F_2$$. Neglect friction and the mass of the pulleys and cords.

The net force acting on the object by the ropes should be $$4F_1 - 1F_2$$ according to the solution manual. I guess I never fully understood how different tensions act on objects. I understand intutitvely that there should be only $$1F_2$$ force acting on the box on the left because the string pulls on the box with exactly $$1F_2$$ force.

But for the right portion of the box diagram where $$F_1$$ is acting on the object. Why is it $$4F_1$$ exactly? I am just trying to intuitively make sense out of the equation as opposed to just counting ropes. This may be a real simple question but please help me out.

Edit: I am not asking for a hw solution (I have access to the solution manual), I am simply asking for a deeper explanation on how tensions act as forces in this diagram to solidify my understanding.

• Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this page in the site help for more on what topics you can ask about here. – John Rennie Mar 27 '19 at 5:19
• I disagree. The OP has the solution but wants help in understanding a principle of Physics. Please read the post before engaging in moderator clamp down. – Paul Childs Mar 27 '19 at 6:42

This is a nice example of a block and tackle system, where two mechanical "flavours" are realized at the same time: The "rove to advantage" on the right, i.e. the pull on the rope is in the same direction in which the load is to be moved and "rove to disadvantage" on the left, i.e. the pull opposes the direction in which the load is to be moved. This is the reason why "counting ropes" as you called it will not lead you to the solution here.

In general, a block and tackle system allows you to move heavy weight (represented by a Force $$F$$) by increasing the distance $$s$$ that you have to cover. The work $$W = F \cdot s$$ always stays the same.

Check the example 1 from Wikipedia below: If you pull the string on the left, you move the weight up. Due to the pulley configuration however, the weight will move half the distance of what you pulled. The necessary force is also 0.5 times compared to simply lifting the weight.

Now check the example 2 from Wikipedia: You have the same amount of ropes and the same amount of pulleys. However, they are attached in a different way: The weight is fixed at two points of the pulley system. If you do the maths, you will find that you only have to exert one third of the force and pull 3 times the distance compared to simply lifting the weight.

Example 1 corresponds to the left side of your system. Force is transferred in a 1:1 ratio to the object. Example 2 corresponds to the right side. Here, force is transferred in a 4:1 ratio. I recommend the wiki article to solidify your understanding. If the maths is unclear, comment below.

In simple mechanics problems like this, where the mass of the rope and the pulleys and friction on the pulleys can be neglected, then you can assume there is the same tension $$T$$ throughout the whole length of the rope.

The rope is attached to the box (via pulleys) at four points on the right hand side of the block, so the force exerted by the rope on the right hand side of the block is $$4T$$. And because the force acting on one end of the rope is $$F_1$$ then you know $$T=F_1$$. This is where the $$4F_1$$ term comes from.