Solve Sliding Box Problem: Force Normal, Constant Velocity & Acceleration

[KatieLynn]

Homework Statement

A student is pulling a 25 kg trunk up an incline that has an angle with the horizontal of 20 degrees. She has a strap tied to the trunk which is parallel to the surface of the incline. If the coefficient of kinetic friction between the incline and the trunk is 0.5 fine
a)Force Normal
b) the force necessary to keep the trunk moving at a constant velocity.
c)the force she would have to apply to accelerate the trunk up the incline at 0.5 m/s squared.

Homework Equations

f(net)=ma
coeff.=(force friction)/(force normal)

The Attempt at a Solution

So I think I found (A) to be force normal = 230N by doing Fn=(m)(g)(cos20)
I tried doing this for B (25)(-9.81)(sin20)-(0.5)(25)(-9.81)(cos20) + ForceX =0
I got 31 but that's not right...

 

[Astronuc]

(25)(-9.81)(sin20) + (0.5)(25)(-9.81)(cos20) + ForceX = 0

The friction force opposes motion, so it must point down the ramp, and the weight component parallel to the ramp must also point down the ramp. So these two force must have the same sign (direction). Both oppose the pulling force up the ramp.

 

[ogb p]

First, let's clarify the given information:
- The trunk has a mass of 25 kg.
- The incline has an angle of 20 degrees with the horizontal.
- The coefficient of kinetic friction between the incline and the trunk is 0.5.
- The student is pulling the trunk up the incline with a strap parallel to the surface.

a) To find the force normal (Fn), we can use the equation Fn = mgcosθ, where m is the mass of the trunk, g is the acceleration due to gravity (9.81 m/s^2), and θ is the angle of the incline. Plugging in the values, we get Fn = (25 kg)(9.81 m/s^2)cos(20 degrees) = 230.3 N.

b) To find the force necessary to keep the trunk moving at a constant velocity, we need to consider the forces acting on the trunk in the horizontal direction. These forces are the force of gravity (mg sinθ) and the force of kinetic friction (μFn), where μ is the coefficient of kinetic friction. Since the trunk is moving at a constant velocity, the net force in the horizontal direction is zero. This means that the force of gravity and the force of kinetic friction must cancel each other out. Therefore, we can set up the equation mg sinθ = μFn. Plugging in the values, we get (25 kg)(9.81 m/s^2)sin(20 degrees) = (0.5)(230.3 N) = 115.1 N.

c) To find the force necessary to accelerate the trunk up the incline at 0.5 m/s^2, we can use the equation F(net) = ma, where F(net) is the net force acting on the trunk, m is the mass of the trunk, and a is the acceleration. Since we want to accelerate the trunk up the incline, the net force must be in the same direction as the acceleration. Therefore, we can set up the equation F(net) = mg sinθ + μFn. Plugging in the values, we get F(net) = (25 kg)(9.81 m/s^2)sin(20 degrees) + (0.5)(230.3 N) = 115.1 N + 115.1 N = 230.2 N. This is the force the student would have to