Minimum Force for Sliding Up Ramp | Ramp Problem Homework Help

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To determine the minimum horizontal force required to slide a 4.0kg block up a ramp inclined at 37 degrees with a static friction coefficient of 0.23, the normal force (N) and frictional force (Ff) must be calculated accurately. The correct normal force is 59.4N, which differs from the initially calculated 31.3N. The applied force (F) needed to overcome both friction and gravity is 46.6N, indicating that the initial calculations for both N and F were incorrect. The discussion emphasizes the importance of including the y-component of the applied force in the calculations. Accurate determination of these forces is crucial for solving the problem effectively.
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Homework Statement


What is the minimum horizontal force that must be applied to the stationary 4.0kg block so that the block will start to slide up the ramp? the ramp is inclined at a 37degrees angle and the coefficient of static friction between the block and the ramp is 0.23.

Homework Equations


Ff = μN
---

The Attempt at a Solution


y-direction
0 = N - mgcosθ
N = mgcosθ
N = 4(9.8)cos37 = 31.3
------
x-direction
0 = Fcos(θ) - Ff − mgsinθ
F = (mgsinθ + Ff)/cos(θ)
F = [39.2sin(37) + (.23(39.2)(cos37)]/cos37
F = 38.6
==================================================
I have the wrong answer because the correct answer is
N = 59.4N, F = 46.6
Also, i think the way i use to solve F is right but i believe my N is wrong because when i plug in the N of the correct answer i get the same F as the result. Hope you can help me solve this problem thanks a lot
 
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Hi Toila,

Toila said:

Homework Statement


What is the minimum horizontal force that must be applied to the stationary 4.0kg block so that the block will start to slide up the ramp? the ramp is inclined at a 37degrees angle and the coefficient of static friction between the block and the ramp is 0.23.

Homework Equations


Ff = μN
---

The Attempt at a Solution


y-direction
0 = N - mgcosθ
N = mgcosθ
N = 4(9.8)cos37 = 31.3
------
x-direction
0 = Fcos(θ) - Ff − mgsinθ

Notice here that you have the x-component of the applied force. But you have not taken into account the y-component of the applied force anywhere.

F = (mgsinθ + Ff)/cos(θ)
F = [39.2sin(37) + (.23(39.2)(cos37)]/cos37
F = 38.6
==================================================
I have the wrong answer because the correct answer is
N = 59.4N, F = 46.6
Also, i think the way i use to solve F is right but i believe my N is wrong because when i plug in the N of the correct answer i get the same F as the result. Hope you can help me solve this problem thanks a lot
 
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