A worker wants to turn over a uniform, 1250-N, rectangular crate by pulling at 53.0° on one of its vertical sides (?Fig. P11.65?). The floor is rough enough to pre-vent the crate from slipping. (a) What pull is needed to just start the crate to tip? (b) How hard does the floor push upward on the crate? (c) Find the friction force on the crate. (d) What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

Solution 69P We first need to calculate the pulling force on the crate. Then we can proceed to find the normal force exerted by the floor, the friction force and the coefficient of kinetic friction. Let us have a look at the following figure to understand the situation better. Let F be the pulling force. Now, the moment of the y-component of the pulling force has to be equal to the moment of the crate’s weight in order to tip the crate. 0 2.20 (a) Therefore, Fsin 53.0 × 1.50 = 1250 N × 2 m 0 Fsin 53 = 1375 N.m 1375 F = 1.50×sin 53.0 F 1150 N Therefore, the required pull is 1150...