Question

In: Physics

(a) A block of mass m = 3.90 kg is suspended as shown in the diagram...

(a) A block of mass m = 3.90 kg is suspended as shown in the diagram below. Assume the pulley to be frictionless and the mass of the strings to be negligible. If the system is in equilibrium, what will be the reading of the spring scale in newtons? N

(b) Two blocks each of mass m = 3.90 kg are connected as shown in the diagram below. Assume the pulley to be frictionless and the mass of the strings to be negligible. If the system is in equilibrium, what will be the reading of the spring scale in newtons? N

(c) A block of mass m = 3.90 kg is in equilibrium on an incline plane of angle θ = 28.0° when connected as shown in the diagram below. Assume the mass of the strings to be negligible. If the system is in equilibrium, what will be the reading of the spring scale in newtons? N

Expert Solution

(b) Two blocks each of mass m = 3.93 kg are connected as shown in the diagram below.

If the system is in equilibrium, then the reading of the spring scale which will be given as :

we know that, F_N = m g

where, g = acceleration due to gravity = 9.8 m/s²

F_N = 2 (3.9 kg) (9.8 m/s²)

F_N = 76.4 N

(c) A block of mass m = 3.9 kg is in equilibrium on an incline plane of angle, = 28⁰ when connected as shown in the diagram below.

If the system is in equilibrium, then the reading of the spring scale which will be given as :

using an equation, we have

F_N = mg cos

where, m = mass of the block = 3.9 kg

= inclination angle = 28 degree

then, we have

F_N = (3.9 kg) (9.8 m/s²) cos 28⁰

F_N = (38.2 N) (0.8829)

F_N = 33.7 N

genius_generous answered 9 months ago

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