Hey all, i don't like asking 4 help like this, but i have this really bad teacher who never tells us how to do anything, and just gives us questions to do, if we can't do them she goes mental and gives us detention. So if any one can help with these questions it would be so much appreciated!
5. A fork-lift truck is raising a container of car batteries with an acceleration of 1.2ms-2. The normal contact force on the container from the horizontal forks is 1485N. Calculate the mass of the load.
7. The pilot of a hot air balloon has mass 85kg. As the balloon leaves the ground, the normal contact force on the pilot from the floor of the balloon immediately increases to 901N and remains at this value for the first stage of the ascent. Calculate the acceleration of the balloon in this stage of its motion.
9. A lift starting from rest moves downwards with constant acceleration. It covers a distance s in time t , where s= 1/6 gt squared. A box of mass m is on the floor in the lift. Find, in terms of m and g, an expression for the normal contact force on the box from the lift floor.
11. A man of mass M kg and his son of mass m kg are standing in a lift. When the lift is accelerating upwards with magnitude 1.2ms-2 the magnitude of the normal contact force exerted on the man by the lift floor is 880N. When the lift is moving with constant speed the combined magnitude of the normal contact forces exerted on the man and the boy by the lift floor is 980N. Find the values of M and m.
If you are able to do any of these, then it means i will dare go back to school. Thankyou all soooooooooo much!!!
right, ive just done m1 too but i havent really revised it yet (after the holidays yeh?) so im not sure if these are right
5. Resultant force = mass x accel
1485 - mg = m(1.2)
1485 = m(1.2) + mg
1485 / (1.2 + g) = m
m = 135Kg
7. again using the same rule
901 - 85g = 85a
a = 901/85 - g
a = 0.8ms^-2
9. not sure about tbh but i think this is it
s = 1/6gt^2
you have u = 0
t = t
a = ?
s = ut + 1/2at^2
1/6gt^2 = 0 + 1/2at^2
divide by 1/2
1/3g = a
sub this back into the lift
R - m = 1/3g(m)
R = 1/3gm + m
R = m(1/3g + 1)
11. use the same way as 7., i think you need to deal with when the lift is accelerating and when its moving constantly (acceleration = 0). you may be able to take the components individually (ie man and child separaretly). probably.
(accelerating downwards : 880 - Mg = 1.2M
880 / (1.2 + g) = M = 80Kg
constant accel : 980 - 80g - mg = 0
980/g - 80 = m = 20Kg)