Laws of motion/static friction problem

[Leesh09]

Homework Statement

Two astronauts are unloading scientific equipment from the spaceship that has just landed on the Moon surface. To prevent a box from sliding down an inclined ramp, astronaut A pushes on the box in the direction parallel to the incline, just hard enough to hold the box stationary. In an identical situation astronaut B pushes on the box horizontally. Regard as known the mass m of the box, the coefficient of static friction s between box and incline, and the inclination angle . (a) Determine the force A has to exert. (b) Determine the force B has to exert. (c) If m = 2.00 kg, = 25.0°, and s = 0.160, who has the easier job? (d) What if s = 0.380? Whose job is easier? The acceleration of gravity on the Moon is 1.625 m/s2.

Homework Equations

fstatic= coefficient of friction * normal force

The Attempt at a Solution

For A, the applied force is on the x plane so the forces in play are the applied force, the static friction force, and the x component of the normal force/mg. Both the static friction force and applied force are working against gravitational force to keep the box in place and there is no movement, so mg*sin theta = applied force + static friction force so the applied force= mg*sin theta - (coefficient of friction)(gravitational constant)??
I don't know how to tackle B though, since the applied force is neither parallel nor perpendicular to the incline.

 

[rl.bhat]

In the first case what is the frictional force?
The applied force is in the upward direction along the inclined plane. What is the direction of the frictional force and mgsinθ?
In the case of B, resolve the force into components. One along the inclined plane and the other normal to the inclined plane. Then identify the directions of mgsinθ, frictional force and the component of applied force.

 

[tomcue]

I would approach this problem by first identifying the key variables and equations involved. The key variables in this problem are the mass of the box (m), the coefficient of static friction (μs), the inclination angle (θ), and the acceleration of gravity on the Moon (g). The key equations involved are the equation for static friction (fstatic= μs * normal force) and Newton's Second Law (F=ma).

Using these equations, we can determine the force that each astronaut needs to exert to keep the box stationary. For astronaut A, the applied force (F) is in the same direction as the static friction force (fstatic), so we can set up the equation as follows: F + fstatic = mg*sinθ. Rearranging this equation, we get F = mg*sinθ - fstatic. Plugging in the given values of m, θ, and μs, we can calculate the force A needs to exert.

For astronaut B, the applied force (F) is not parallel or perpendicular to the incline, so we need to break it down into its components. The x component of the applied force will be counteracted by the static friction force, while the y component will be counteracted by the normal force (mg*cosθ). Setting up the equations, we get Fx = fstatic and Fy = mg*cosθ. Rearranging and substituting in the values, we can calculate the force B needs to exert.

Comparing the two forces, we can see that B needs to exert a greater force than A, making their job more difficult. This is because the force B needs to exert is split into two components, while A only needs to exert one force.

If the coefficient of static friction is increased to 0.380, the static friction force will also increase. This means that both astronauts will need to exert a greater force to keep the box stationary, but B will still need to exert a greater force due to the additional component of the applied force.

In conclusion, astronaut A has the easier job in both scenarios, as they only need to exert one force in the direction parallel to the incline. Astronaut B has a more difficult job as they need to exert two forces in different directions to keep the box from sliding down the incline.