Need Help with Two Blocks & Rope Physics Problem

[Cougar1112]

Ok I am new this forum so hello to all. Here is the problem that I've been working on. Is this right or wrong and what is the right answer?


Two blocks are connected by a rope that wraps around a pulley. A force of 50 N acts on the 5 kg mass at an angle of 36 degrees.

1. what is the normal force on the 5kg mass?(I got 15.55 N)
2. what is the acceleration of the blocks and the tension in the rope?(2.6 m/s squared for acceleration, but I can't figure out the tension)

Im sure I got this question wrong so that's why I am using this board. Also can someone post up the formulas that they used to solve this problem.

THANKYOU!

 

[Doc Al]

Please describe the physical arrangement of the blocks in more detail (a picture would help). Also: Post your work, not just your answers.

 

[quicknote]

Hello and welcome to the forum! It's great that you are seeking help and clarification on this problem. Based on the information provided, here is how I would approach this problem:

1. To find the normal force on the 5kg mass, we can use the formula F = ma, where F is the force, m is the mass, and a is the acceleration. In this case, the force acting on the mass is the weight (mg) and the acceleration is due to gravity (9.8 m/s^2). So the normal force would be 5kg x 9.8 m/s^2 = 49 N. This means that the 50 N force is balanced by the normal force, so the 5kg mass is not moving up or down.

2. To find the acceleration of the blocks and the tension in the rope, we can use the formula F = ma again. This time, the force acting on the blocks is the tension in the rope, and the acceleration is the same for both blocks since they are connected by the rope. So we have the equation T = (5kg + 2kg) x a. We also know that the force of 50 N is acting on the 5kg mass at an angle of 36 degrees, so we need to find the component of this force that is acting in the direction of motion. This can be done by using the formula Fcos(theta), where F is the force and theta is the angle. So in this case, the component of the force acting in the direction of motion is 50 N x cos(36 degrees) = 40.6 N. Therefore, we have the equation T = (5kg + 2kg) x a = 40.6 N. Solving for a, we get a = 40.6 N / 7kg = 5.8 m/s^2. This is the acceleration of both blocks.

To find the tension in the rope, we can plug in the value for the acceleration (5.8 m/s^2) into the equation T = (5kg + 2kg) x a. This gives us T = 7kg x 5.8 m/s^2 = 40.6 N. So the tension in the rope is 40.6 N.

I hope this helps! In terms of formulas, the key ones used here are F = ma and Fcos