Question

3. Determine the maximum force P that can be applied without causing movement of the 300-lb crate that has a center of gravity at G. The coefficient of static friction at the floor is Is = 0.4. (20 Points) 1.5 ft 1.5 ft 2.5 ft P- 'T 4.5 ft 3.5 ft

Answer 1

A) Given that Mg 2011, crote weight = 30026 2.5ft 4.SFE 3002 | 3.5ft 1.5Ft 1.5ft As we are applying force 'p' Hilfe will be some opposite Force (f) at 7 and due to weight there will be normal seaction H = P_F=0 P = MN p=0.41300) 3002b P=120 lb 4.597 let vã ozume the chate slips EMO =O => NX =P(4.5ft) 300 x = 120145 Check (n=1185t) (2 1.5 ft) (XV-0 => N= 3001 Rough free Body diagram .. No slipning ociuto berans> 1.5

Since tipping oclora then correct Free BODY diagram Isstylish Given condition is no movement of yote. Therefore {ry=0 SA 300b P=F 3.5Ft EFV = 0 =>/N=30026 TSFEIN CONSEIL FBD of crate {MO =D 2145FD = N115) 214,5) = 300(5) ?=100 hb P=F=10024] [< AN =12016)