A safety mechanism of a lift is simplified by the figure below. A 300 kg block is to slide over an inclined surface. The coefficient of friction between the block rollers and the inclined surface is u=0.3. If the system is in equilibrium, work out the following: (a) Free body diagram of the complete system. (b) Find the normal force N on the block and the maximum friction force Fmax. (c) Find the mass of the lift m. required to satisfy the equilibrium and prevents the block moving downward, (d) Find tension force in the cable (e) If the mass of the lift is 150 kg, p=0.3, and surface angle is increased to 30° will the block move upward? Explain the answer. 300 kg block 20° Cable Lift mo

A safety mechanism of a lift is simplified by the figure below. A 300 kg block is to slide over an inclined surface. The coefficient of friction between the block rollers and the inclined surface is u=0.3. If the system is in equilibrium, work out the following: (a) Free body diagram of the complete system. (b) Find the normal force N on the block and the maximum friction force Fmax. (c) Find the mass of the lift m. required to satisfy the equilibrium and prevents the block moving downward, (d) Find tension force in the cable (e) If the mass of the lift is 150 kg, p=0.3, and surface angle is increased to 30° will the block move upward? Explain the answer. 300 kg block 20° Cable Lift mo