What is the maximum value the string tension can have before the can slips? The coefficient of static friction between the can and the ground is 0.76.

T= 38.38 N

Explanation:

Here

mass of can = m = 3 kg

g= 9.8 m/sec2

angle θ = 40°

From figure we see the vertical and horizontal component of tension force T

If the can is to slip - then horizontal component of tension force should become equal to force of friction.

First we find force of friction

Fs= μ R

where

μ = 0.76

R = weight of can = mg = 3 × 9.8 = 29.4 N

Now horizontal component of tension

Tx= T cos 40 = T× 0.7660  N

==>T× 0.7660 = 29.4

==> T= 38.38 N

for this one it would be d bump/forearm pass

50 inches should be the resultant vector