Block A, with a mass of 50 kg, rests on a horizontal table top. The coefficient of static friction is 0.40. A horizontal string is attached to A and passes over a massless, ideal pulley as shown.

What is the smallest mass of Block B that will start A moving when it is attached to the other end of the string?

A person has a choice while trying to move a crate across a horizontal pad of concrete: push it at a downward angle of 30 degrees, or pull it at an upward angle of 30 degrees.

a) Which choice is most likely to require less force on the part of the person?
1. pushing at a downward angle
2. pulling at the same angle, but upward
3. pushing or pulling shouldn't matter

The answer I chose was 2 because this would decrease friction resulting in decrease of normal force.

b) If the crate has a mass of 50.0 kg and the coefficient of friction between it and the concrete is 0.750, calculate the force required to move it across the concrete at a constant speed in both situations.

This is what I don't really get. I was thinking of using Fpcos(30) = uN with N being (mg + Fpsin(30)) for pushing and then Fpcos(30) = uN with N being (mg - Fpsin(30)) for pulling. Any ideas?

Block B pulls on A

so at the point when weight of B equals the friction on A..

massB*g=massA*mu*g

On the crate, your approach is correct.

upward angle
Fcos30=.75(50g-Fsin30)
f(cos30+.75sin30)=.75*50g
solve for F, and for the downward angle
f(cos30-.75sin30)=.75*50g

To find the smallest mass of Block B that will start Block A moving, we need to consider the force of friction acting on Block A. The force of friction can be determined using the coefficient of static friction. Here's how you can calculate it step by step:

1. Identify the forces acting on Block A:
- The force of gravity acting downward with a magnitude of mg, where m is the mass of Block A (50 kg) and g is the acceleration due to gravity (9.8 m/s^2).
- The normal force exerted by the table, which will be equal in magnitude but opposite in direction to the force of gravity.

2. Calculate the normal force: Since Block A is not accelerating vertically, the normal force must balance the force of gravity. Therefore, the normal force is equal in magnitude to the force of gravity, which is mg.

3. Determine the maximum force of static friction: The maximum force of static friction can be found by multiplying the coefficient of static friction (0.40) by the normal force. So, the maximum force of static friction is μs * mg.

4. Calculate the tension in the string: Since the string is massless and ideal, the tension is the same throughout the string. Therefore, the tension in the string is equal to the force of static friction.

5. Use the tension in the string to find the mass of Block B: The tension in the string must be greater than or equal to the static friction for Block A to start moving. We can equate the tension in the string to the gravitational force acting on Block B, which is mB * g, where mB is the mass of Block B. So, μs * mg = mB * g.

6. Solve for the mass of Block B: Divide both sides of the equation by g to isolate the mass of Block B. So, μs * m = mB. Now, substitute the given value of the coefficient of static friction (0.40) and the mass of Block A (50 kg) into the equation to find the mass of Block B.

mB = 0.40 * 50 kg
mB = 20 kg

Therefore, the smallest mass of Block B required to start Block A moving when it is attached to the other end of the string is 20 kg.