How Do You Calculate Forces on Inclined Planes?

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The discussion focuses on calculating forces on inclined planes through two specific problems involving tension, normal force, and acceleration. For the first problem, the user calculates the tension in the cord and the normal force acting on an 8.2 kg block on a 37° incline, but struggles with the acceleration calculation after cutting the cord. In the second problem, a 104 kg crate is pushed up a frictionless ramp at a 29° angle, where the user correctly finds the horizontal force needed but encounters issues with calculating the normal force. Key points include the importance of resolving gravitational forces into components and applying Newton's second law. The conversation highlights the need for clarity in force calculations on inclined planes.
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Homework Statement



I have two questions:

1. Let the mass of the block be 8.2 kg and the angle θ be 37°. Find (a) the tension in the cord and (b) the normal force acting on the block. (c) If the cord is cut, find the magnitude of the block's acceleration.

2.In the figure, a crate of mass m = 104 kg is pushed at a constant speed up a frictionless ramp (θ = 29°) by a horizontal force f-> . The positive direction of an x-axis is up the ramp, and the positive direction of a y-axis is perpendicular to the ramp. (a) What is the magnitude of f-> ? (b) What is the magnitude of the normal force on the crate?



Homework Equations



F_net=ma
Maybe T-m*g*sin(theta)=ma


The Attempt at a Solution



Ok, for 1.a. I used 8.2kg*9.8m/s^2*cos(29)=48.3619N to find the tension on the cord. For b I used 8.2*9.8*sin(29)=64.1783N. For 3. I am having trouble because my T and mgsintheta are the same.

For 2a. I used 104kg*9.8m/s^2tan(29) to get 564.92N and for B is got 1019N*cos(29) but the computer says this is wrong. The only one I got right was 2.a, 1.a and 1.b were revised so they might be right.



Can anyone help? Thank you very much
 
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Welcome to PF.

What is the diagram of the problems you are solving.
 
Last edited:
1)The force component of gravity pointing down the ramp is sinθ.

Hence the Tension to off set that in the direction up the ramp is m*g*sinθ = m*a where a is the answer to your part c.

The Cosθ component of g determines the force normal to the incline.

2) For part b you have two components of force to account for.

From the F you have F*sinθ
From mass and gravity you have m*g*Cosθ
 
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