Newton's Laws of motion problem

[sepah50]

Homework Statement

A block weighing 70.0 N rests on a plane inclined at 25.0° to the horizontal. A force F is applied to the object at 35.0° to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.333 and 0.156.

What is the minimum value of F that will prevent the block from slipping down the plane?

Homework Equations

$$\sum$$F_y= Sum of all Forces = Newton's Second Law

The Attempt at a Solution

So in the beginning I draw the Angles and make a Free Body Diagram. I sum up all the forces which is
$$\sum$$F_y = $$\eta$$ + Fsin(10) - 70cos(25) = ma_y

ma_y is 0 since block is not moving away from the plane

what I get then is $$\eta$$= .173648F - 63.4415

After this point I know I have to find F_smax which equals $$\mu$$_s$$\eta$$. After this part I get lost but I can't find $$\eta$$$$\mu$$_s
From what I remember $$\mu$$$$\eta$$_s = mg*sin$$\theta$$. Hope somebody can help!

 

[alphysicist]

Hi sepah50,

Homework Statement

A block weighing 70.0 N rests on a plane inclined at 25.0° to the horizontal. A force F is applied to the object at 35.0° to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.333 and 0.156.

What is the minimum value of F that will prevent the block from slipping down the plane?

Homework Equations

$$\sum$$F_y= Sum of all Forces = Newton's Second Law

The Attempt at a Solution

So in the beginning I draw the Angles and make a Free Body Diagram. I sum up all the forces which is
$$\sum$$F_y = $$\eta$$ + Fsin(10) - 70cos(25) = ma_y

ma_y is 0 since block is not moving away from the plane

what I get then is $$\eta$$= .173648F - 63.4415

I think you have a couple of sign errors here.

After this point I know I have to find F_smax which equals $$\mu$$_s$$\eta$$. After this part I get lost but I can't find $$\eta$$$$\mu$$_s

When you summed up all the forces, you did it only in the y-direction (perpendicular to the plane). What is the sum of the forces in the x-direction (parallel to the plane)?

From what I remember $$\mu$$$$\eta$$_s = mg*sin$$\theta$$. Hope somebody can help!

 

[sepah50]

Thanks! :)

 

[cgrbym]

thanks

 

[rwisz]

First, we need to determine the normal force (F_N) acting on the block, which is equal to the component of the weight of the block perpendicular to the plane. We can find this using trigonometry: F_N = 70.0 N * cos(25.0°) = 63.4415 N.

Next, we need to find the maximum possible frictional force (F_smax) that can act on the block without causing it to slip down the plane. This is given by F_smax = \mu_s * F_N, where \mu_s is the coefficient of static friction. Plugging in the values, we get F_smax = 0.333 * 63.4415 N = 21.1465 N.

Now, we can set up an inequality to find the minimum value of F that will prevent the block from slipping down the plane: F * sin(35.0°) ≥ F_smax. Solving for F, we get F ≥ F_smax / sin(35.0°) = 21.1465 N / sin(35.0°) = 38.8591 N.

Therefore, the minimum value of F that will prevent the block from slipping down the plane is 38.8591 N.