Minimum Force for Sliding Up Ramp | Ramp Problem Homework Help

[Toila]

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

 

[alphysicist]

Hi Toila,

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

 

[kerimek]

I can help you understand the correct approach to solving this problem. First, we need to understand the forces acting on the block. There is the weight of the block, mg, acting downwards, and the normal force, N, acting perpendicular to the surface of the ramp. There is also the frictional force, Ff, acting parallel to the surface of the ramp.

In order for the block to start sliding up the ramp, the applied horizontal force, F, must overcome the static frictional force, Ff. This means that F must be greater than or equal to Ff.

Using the equation Ff = μN, we can determine the maximum possible value for F. In this case, μ = 0.23 and N = mgcosθ. So, Ff = 0.23(4)(9.8)cos37 = 9.8N. This means that the minimum force required to start sliding up the ramp is 9.8N.

However, we also need to consider the component of the weight of the block that is acting parallel to the surface of the ramp, mg sinθ. This component will decrease the required minimum force, as it acts in the opposite direction to Ff. So, we can modify our equation to be F ≥ (mg sinθ + Ff)/cosθ. Plugging in the values, we get F ≥ (4)(9.8)sin37 + 9.8/cos37 = 46.6N.

Therefore, the minimum horizontal force that must be applied to the stationary 4.0kg block is 46.6N. This value is higher than the one you calculated because you did not take into account the component of the weight acting parallel to the ramp. I hope this helps clarify the problem for you.