A block of mass m1 (labeled by U) is placed on top of a larger block of mass m2 (labeled by L). A man is trying to pull the blocks to the right via a spring (labeled by S) attached to the large block, but the blocks do not move. The friction force between the large block and the ground is FL,G. What is the magnitude of the friction force between the large and small block?

https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys100/fall10/preflights/12/fig24_fa05.gif

0
(m1/m2)FL,G
(m1/(m1 + m2))FL,G how to start?



A truck of mass M is moving with a constant acceleration a up a hill that is inclined at an angle α with respect to the horizontal.

https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys100/fall10/preflights/12/pf5_2.gif

In case I, the truck carries a box of mass m and in Case II, the mass of the box is 2m. In both cases the coefficient of static friction between the box and the bed of the truck is μS. Compare aI, the maximum acceleration the truck can have before the box slips in Case I, to aII, the maximum acceleration the truck can have before the box slips in Case II.
aI < aII
aI = aII
aI > aII again, how can I start?

1. I'm assuming that the coefficient of static friction is the same between the blocks and the ground.

The friction on between the large block and the ground is given by:

F_L = \mu R_1

Find R

What is the equation for the frictional force between the small block and the large block?
F_S = \mu R_2
You'll need the normal force on the small block.

Substitute the coefficient of friction now, from the first equation into the second equation.

2. Find the maximum frictional force that can act on the box, that is when the frictional force is equal in magnitude to the component of gravity of the box.

The one which can produce a greater frictional force can be accelerated to a greater extend by the truck.