# 3 Forces Questions

Can anyone help me with these motion questions?

1. Find the angle of a supermarket ramp that will alow a person to push a 30 kg shopping cart. The people can push with 50 N of force and friction can be accounted for with a coefficient of 0.10.

For this question I broke up gravity in the x and y direction since it is on a ramp. Fg in y=mgcos(Theta) Fg in x=mgSin(Theta)

My two Fnet equations are:
Up is positive Fnet=Fn-Fgy
0=Fn-Fgy
Fn=mgCos(Theta)

Right is positive Fnet=Fapplied-Ffriction-Fgx
0=50-(0.1)*(30)*(9.8)*cos(Theta)+(30)*(9.8)*sin(Theta)
50=29.4cos(Theta)+294sin(Theta)

And then I dont know what to do next.

2. Block m1 has a mass of 4 kg and block m2 has a mass of 2 kg. The coefficient of friction between m2 and the horizontal surface is 0.50 and the coefficent between m1 nd the inclined surface is 0.289. The incline is 30(degrees). Find the tension in the string and the acceleration of the blocks.

The answer I got is Tension=29.4 N and acceleration is 9.8 m/s^2. Am i right?

3. Your friend has been chained on the wall of a dungeon. You can not get into the room, and the only way to save him is to pull on a rope and drag the 100 kg box with the food and keys on it. Your friend can only reach 1 m and the top of the box is 1 m above the floor and 7 m away from the wall where he is chained. The rope goes through a wall 5 m above the floor. The coefficient of friction between the box and the florr is 0.5. THe rope is old and will break if you pull with more than 550 N. Can you save your friend? If you cannot save him how cloase does the rope get before the rope broke?

HallsofIvy
Homework Helper
. Find the angle of a supermarket ramp that will alow a person to push a 30 kg shopping cart. The people can push with 50 N of force and friction can be accounted for with a coefficient of 0.10.

For this question I broke up gravity in the x and y direction since it is on a ramp. Fg in y=mgcos(Theta) Fg in x=mgSin(Theta)

My two Fnet equations are:
Up is positive Fnet=Fn-Fgy
0=Fn-Fgy
Fn=mgCos(Theta)

Right is positive Fnet=Fapplied-Ffriction-Fgx
0=50-(0.1)*(30)*(9.8)*cos(Theta)+(30)*(9.8)*sin(Theta)
50=29.4cos(Theta)+294sin(Theta)

And then I dont know what to do next.

Well, I got confused as to which direction was x and which y!
If you are taking x to be in the direction of the ramp and y perpendicular to the ramp, then, yes, Fgy= -mg cos(Theta) and Fgx= -mg sin(Theta). Fgy is the force perpendicular to the ramp and that is what is causing the friction: Ffriction= 0.10 Fgy= -0.10mg cos(Theta). The Net force in the x direction (along the ramp) is, then
50- 0.10 mg cos(Theta)- mg sin(Theta): force applied to push the cart, minus the friction force, minus the force of gravity.
We are given that m= 30 kg so mg= (30)(9.8)= 294 Newtons. The net force is 50- (0.10)(294)cos(Theta)- 294 sin(Theta)
= 50- 29.4 cos(Theta)- 294 sin(Theta). In order for someone to be able to push the cart up the ramp, that must be greater than 0. Solve the equation 50- 29.4 cos(Theta)- 294 sin(Theta)= 0 to find the angle Theta at which that becomes true.

2. Block m1 has a mass of 4 kg and block m2 has a mass of 2 kg. The coefficient of friction between m2 and the horizontal surface is 0.50 and the coefficent between m1 nd the inclined surface is 0.289. The incline is 30(degrees). Find the tension in the string and the acceleration of the blocks.

The answer I got is Tension=29.4 N and acceleration is 9.8 m/s^2. Am i right?
Without even doing any calculations, that acceleration can't be right! That's the acceleration of a free falling body and these are weights on an incline with friction. The acceleration must be lower than 9.8 m/s^2.

If I read this right, m1 is on a horizontal surface and m2 on a surface incline 30 degrees to the horizontal. The gravitational force perpendicular to the surface is -mg cos(Theta) so the frictional force is mg cos(Theta)(0.289)= 4(9.8)(cos(30))(0.289)= 9.8 Newtons. The gravitational force in the direction of the surface is -mg sin(Theta)= -4(9.8)sin(30)= -19.6 Newtons.

The gravitational force on block m1, on the horizontal surface is -mg= -2(9.8)= -19.6 Newtons so the frictional force on it is 19.6(0.5)= 9.8 Newtons.

The net force on both blocks is the gravitational force down the incline minus the two friction forces: 19.6- 9.8- 9.8= 0. The two blocks won't move! If there were only the block m2 on the incline, the force calculation would be 19.6- 9.8= 9.8 Newtons so that it would move. Since the upper block is keeping it from moving, the upper block2 must be pulling back on it with force of 9.8 Newtons: the tension in the string is 9.8 Newtons.

3. Your friend has been chained on the wall of a dungeon. You can not get into the room, and the only way to save him is to pull on a rope and drag the 100 kg box with the food and keys on it. Your friend can only reach 1 m and the top of the box is 1 m above the floor and 7 m away from the wall where he is chained. The rope goes through a wall 5 m above the floor. The coefficient of friction between the box and the florr is 0.5. THe rope is old and will break if you pull with more than 550 N. Can you save your friend? If you cannot save him how cloase does the rope get before the rope broke?

I assume, though the problem doesn't say it, that the rope is place so you can pull the box directly toward your friend. I am also going to assume that the rope is attached to the top of the box. You need to pull the box within 1 meter of the wall.

Let "x" be the distance from the box to the wall. The vertical distance from the top of the box to the "hole in the wall" through which you are pulling the rope is 5-1 = 4 m. The point of that is that the components of the force vector pulling the box are the horizontal and vertical components of a vector along the hypotenuse of that right triangle.. The gravitational vector downward on the box is, of course, -mg= -100(9.8)= -98 Newtons. Since the coefficient of friction is 0.5, the friction force is 0.5(-98)= -49 Newtons. We must apply a horizontal force of at least 49 Newtons to move the box.
Since the components of force on the rope form a right triangle with the same proportions as the right triangle with rope as hypotenuse, we must have 49/x= Fvertical/4 or Fvertical = 4(49)/x= 196/x Newtons.
Since the rope itself is the hypotenuse of that right triangle, we can find the tension in it by the Pythagorean theorem: &radic;(492+ (196)2/x2) That cannot be more than 550 Newtons. The critical point is when &radic;(492+ (196)2/x2)= 550.

Squaring both sides, 492+ (196)2/x2= 5502. Solve for x. If x is less than 1, you can save your friend.